The purpose of this page is to guide someone wishing to learn to program using the C programming language, specifically the ANSI Standard of 1989. Each section consists of<\/p>\n\n\n\n
Some sections also include a link to a video demonstrating an application of the topic or a link to another page in which I go into more detail about my design process.<\/p>\n\n\n\n
Note that the practice problems build upon your knowledge. Some of the practice problems assume you worked the earlier problems.<\/p>\n\n\n\n
start here: Introduction<\/a><\/p>\n\n\n\n Now you need to install the necessary software:<\/p>\n\n\n\n start here: Statements<\/a><\/p>\n\n\n\n Now try the following practice problems:<\/p>\n\n\n\n Practice #1:<\/strong> For the first Practice #2: <\/strong>Write a program that stores the values a = 4, b = 13, and c = 6 as variables of type Mistakes in syntax or logic are a part of programming. I have put together a video demonstrating many of the mistakes new (and not so new) C programmers make, with solutions. <\/p>\n\n\n\n There is a catch-22 to this video. If you are new to C programming and watch it now, you won’t understand much of it. If you wait until you have reached the end of this self-learning guide to watch the video, then you have likely already made some of these mistakes. I would encourage you to watch the video early on to at least plant some ideas in your head and then revisit the video as you gain experience.<\/p>\n\n\n\n video: Common C Mistakes<\/a><\/p>\n\n\n\n start here: Conditionals<\/a><\/p>\n\n\n\n then here: Logical Operators<\/a><\/p>\n\n\n\n then here: application of conditionals<\/a><\/p>\n\n\n\n Practice #1:<\/strong> Compare the output from the following programs. Why is it different?<\/p>\n\n\n Practice #2: <\/strong>The following program doesn’t print anything. Why? Using only a set of curly braces, change the program such that the second Practice #3: <\/strong>Write a program that prompts a user for an integer, checks for divisibility by 2 and 3, and then prints the statements given by the truth table below. To prompt the user, you will need the following code:<\/p>\n\n\n start here: Loops<\/a><\/p>\n\n\n\n then here: application of loops<\/a><\/p>\n\n\n\n Practice #1:<\/strong> Rewrite the following program, replacing the Practice #2: <\/strong>You have a program that prompts a user for two integers, which are stored in the variables Practice #3: <\/strong>Write a program that prints the integers from 1 to 100 as a 10×10 table with the values right-aligned within their column. Each row will begin with the character ‘X’. The results will be<\/p>\n\n\n start here: Functions, part 1<\/a><\/p>\n\n\n\n then here: Functions, part 2<\/a><\/p>\n\n\n\n then here: application of functions, part 1<\/a><\/p>\n\n\n\n then here: application of functions, part 2<\/a><\/p>\n\n\n\n Practice #1:<\/strong> Write a function declaration for a function with the name Practice #2:<\/strong> The following program makes a function call to a function called Practice #3:<\/strong> Write a function that when given a two positive integers If the function is called with the code below<\/p>\n\n\n then the output will be<\/p>\n\n\n Practice #4:<\/strong> Write a function that when given an integer x, returns the second smallest divisor of x (not counting 1). For example, if x = 12, then the function would return 3 since the divisors are 1, 2, 3, 4, 6, and 12 (remember, we are not counting 1). If x does not have a second smallest divisor, then the function should return zero.<\/p>\n\n\n\n Practice #5:<\/strong> Write a program that produces the integers in the range of 2 to 100, but only prints those that are prime. If the number produced is not prime, print two periods instead. Everything printed should result in a 10×10 table of either values printed or spaces. The results will be<\/p>\n\n\n Practice #6: Reading code is an invaluable skill as a programmer. You need to be able to read your own code as well as others.<\/p>\n\n\n\n What does this program print? Try constructing a table and filling it in as you process the loop. Your table might start like this:<\/p>\n\n\n start here: One-Dimensional Arrays<\/a><\/p>\n\n\n\n then here: Multidimensional Arrays<\/a><\/p>\n\n\n\n then here: application of arrays, part 1<\/a><\/p>\n\n\n\n design discussion: Simulating a Perfect Shuffle<\/a><\/p>\n\n\n\n Practice #1:<\/strong> Write a program that stores the numbers 1, 2, 7, 18, 23, and 40 in an array. Then move through the array, producing the sum of the even integers.<\/p>\n\n\n\n Practice #2:<\/strong> For the program below, write the code that will store the values in Practice #<\/strong>3:<\/strong> Modify the program below to swap the elements between the two arrays. That is, after you are done then array1[0] = 99 and array2[0] = 1 and so on for each corresponding array element.<\/p>\n\n\n Practice #<\/strong>4:<\/strong> Modify the looping structure of the program below so that it produces the output shown. Hint: this can be done by changing only two lines.<\/p>\n\n\n Practice #5:<\/strong> Write a program that declares a 2D array with three columns. The program passes the array to a function that replaces each element in the center column with the sum of the other two elements on the same row. For example, if the first row of the array is<\/p>\n\n\n then after the call to your function this row would be<\/p>\n\n\n Note that this can be done with a single loop that generates the subscript of the row. Also create a second function that prints a 2D array with three columns.<\/p>\n\n\n\n Practice #6:<\/strong> Finish the program below by writing the code to change any number in the array to 99 if, and only if, the values immediately before and after the current number are both even. The first and last elements of the array should not be changed. The array should be processed beginning with the element at index 1. Note that when processing a number, if the value before it has been changed to 99 that is what should be considered, not the value in the initial version of the array. Your program should still work if I changed the array values.<\/p>\n\n\n Practice #7:<\/strong> Write a function that when given a 2D array with three columns, prints the numbers in each row as a sum and then prints the actual sum of the row values. For example, given<\/p>\n\n\n Practice #8:<\/strong> The program below calls the function start here: Bits, Bytes, and Numbers Bases<\/a><\/p>\n\n\n\n then here: Integer Variable Types<\/a><\/p>\n\n\n\n then here: Floating-Point Variable Types<\/a><\/p>\n\n\n\n Practice #1:<\/strong> or the following program, which of the following is true? If you choose one of the first three answer, explain what the problem is and the solution to it.<\/p>\n\n\n\n Practice #2: <\/strong>Run the following program. Is the output what you expected?<\/p>\n\n\n Practice #3:<\/strong> Why does the number of bytes reported from f() differs from in main()?<\/p>\n\n\n start here: Introduction to Pointers<\/a><\/p>\n\n\n\n then here: Pointers and Arrays, part 1<\/a><\/p>\n\n\n\n then here: Pointers and Arrays, part 2<\/a><\/p>\n\n\n\n Pointers is difficult, yet very important, topic in C. I have put together a separate page that goes into a lot of detail: Pointers<\/a> <\/p>\n\n\n\n Practice #1:<\/strong> What format specifiers (in order) should be used in the printf() statement in the following program? Note that in the program the correct format specifiers have been replaced by Z.<\/p>\n\n\n Practice #2:<\/strong> We have the following code snippet<\/p>\n\n\n What value does each of the following represent? If the answer is an address, state such and include which variable the address is for.<\/p>\n\n\n\n Practice #3:<\/strong> On a machine in which addresses are 8 bytes, what is printed by the following program:<\/p>\n\n\n Practice #4:<\/strong> Rewrite the following program, replacing the subscript notation with the corresponding pointer notation. Only change the parts of the program necessary for the replacement.<\/p>\n\n\n Practice #5:<\/strong> What does the following program print? Creating a table for the values of i and ptr will help.<\/p>\n\n\n Practice #6:<\/strong> Rewrite the following program such that the function has a return type of Practice #7:<\/strong> In this program is a function call to a function called Practice #8:<\/strong> The following program prints out the elements of the array. Why does this work since there is no row offset in the expression in the Practice #9:<\/strong> Write the function change() such that it changes the value of x from 5 to 99.<\/p>\n\n\n start here: Introduction to Strings<\/a><\/p>\n\n\n\n
\n\n\n\nStatements<\/h3>\n\n\n\n
printf()<\/code> has a very large number of format specifiers. There are a number of options that can be used with the format specifiers for altering the way that information is displayed. Run the following program with different values for the format specifiers (sorry about using loops before officially discussing them, but it makes this much simpler).<\/p>\n\n\n\nprintf()<\/code>, some values to try are %2d and %4d. For the second printf()<\/code>, some values to try are %2lf, %4.2lf, and %8.5lf (those are the letters ‘L’ and ‘F’ in lowercase). <\/p>\n\n\n\n#include <stdio.h>\n \nint main(void){\n int i;\n double x;\n \n for(i = 1; i < 10000; i *= 10)\n printf("%d\\n", i);\n \n for(x = 1.234; x < 10000; x *= 10)\n printf("%lf\\n", x);\n}\n<\/pre><\/div>\n\n\nint<\/code> and produces d = (a + b + c)\/5. Then print d to 2 decimal places.<\/p>\n\n\n\n
\n\n\n\nCommon Mistakes<\/h3>\n\n\n\n
\n\n\n\nConditionals<\/h3>\n\n\n\n
\n\/* version 1*\/\n#include <stdio.h>\n \nint main(void) {\n int x = 8;\n \n if(x < 6)\n printf("%d < 6\\n", x);\n else if(x < 8)\n printf("%d < 8\\n", x);\n else if(x < 10)\n printf("%d < 10\\n", x);\n else if(x < 12)\n printf("%d < 12\\n", x);\n else if(x < 14)\n printf("%d < 14\\n", x);\n}\n<\/pre><\/div>\n\n\n\/* version 2 *\/\n#include <stdio.h>\n \nint main(void) {\n int x = 8;\n \n if(x < 6)\n printf("%d < 6\\n", x);\n if(x < 8)\n printf("%d < 8\\n", x);\n if(x < 10)\n printf("%d < 10\\n", x);\n if(x < 12)\n printf("%d < 12\\n", x);\n if(x < 14)\n printf("%d < 14\\n", x);\n}\n<\/pre><\/div>\n\n\nprintf()<\/code> statement will print if age is less than or equal to 50.<\/p>\n\n\n\n#include <stdio.h>\n \nint main(void) {\n int age = 40;\n \n if(age > 50)\n if(age < 75)\n printf("first printf\\n");\n else\n printf("second printf\\n");\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n \nint main(void) {\n int x;\n \n printf("enter an integer: ");\n scanf("%d", &x);\n \n \/* put your logic here *\/\n}\n<\/pre><\/div>\n\n\ndivisible divisible\n by 2 by 3 prints\n---------------------------------------------------\n True True both 2 and 3 are divisors\n True False only 2 is a divisor\n False True only 3 is a divisor\n False False neither 2 nor 3 is a divisor\n<\/pre><\/div>\n\n\n
\n\n\n\nLoops<\/h3>\n\n\n\n
for<\/code> loop with a while<\/code> loop.<\/p>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n int i;\n\n for(i = 8; i > 5; i--)\n printf("%d\\n", i);\n}\n<\/pre><\/div>\n\n\nstart<\/code> and stop<\/code>. Write the code that will produce an integer x<\/code> in the range of start<\/code> to stop<\/code>, inclusive. If x<\/code> is even then your program will print “x is Even”. Otherwise, print the value of x<\/code>. You can assume that the starting integer is less than or equal to the stopping integer.<\/p>\n\n\n\n\nX 1 2 3 4 5 6 7 8 9 10 \nX 11 12 13 14 15 16 17 18 19 20 \nX 21 22 23 24 25 26 27 28 29 30 \nX 31 32 33 34 35 36 37 38 39 40 \nX 41 42 43 44 45 46 47 48 49 50 \nX 51 52 53 54 55 56 57 58 59 60 \nX 61 62 63 64 65 66 67 68 69 70 \nX 71 72 73 74 75 76 77 78 79 80 \nX 81 82 83 84 85 86 87 88 89 90 \nX 91 92 93 94 95 96 97 98 99 100\n<\/pre><\/div>\n\n\n
\n\n\n\nFunctions<\/h3>\n\n\n\n
calcAge<\/code> that will receive a variable of type int<\/code> and has a return type of float<\/code>.<\/p>\n\n\n\nmultNums<\/code>. multNums<\/code> takes two arguments, multiplies them together, and then returns the result. Write the function declaration and function definition for multNums<\/code>. The return type and variable types that you write for multNums<\/code> should match those of the variables involved in the function call of the sample program. Your function definition will include whatever code is necessary to perform the calculation.<\/p>\n\n\n\n#include <stdio.h>\n\n\/* function declaration goes here *\/\n\nint main( void ) {\n int hours = 3;\n float rate = 5;\n double cost;\n\n cost = multNums(hours, rate);\n \n printf("cost is %2.f\\n", cost);\n}\n\n\/* function definition goes here *\/\n<\/pre><\/div>\n\n\na<\/code> and b<\/code> determines how many of the integers in the range of a<\/code> to b<\/code> inclusive are divisible by 3 and how many are divisible by 5. The function will print these two counts as well as how many of the integers in the range are not divisible by 3 or 5. If a number is divisible by 3 and 5, then update the counts for each of 3 and 5. As the function is performing the counts, have it print the matching number. For example, if the current number is 9 then the count for divisibility by 3 will be incremented and you will print that 9 is divisible by 3.<\/p>\n\n\n\n\n#include<stdio.h>\n\nint main( void ) {\n int a = 6, b = 20;\n\n f(a, b);\n}\n<\/pre><\/div>\n\n\n\n6 is divisible by 3\n7 is not divisible by 3 nor 5\n8 is not divisible by 3 nor 5\n9 is divisible by 3\n10 is divisible by 5\n11 is not divisible by 3 nor 5\n12 is divisible by 3\n13 is not divisible by 3 nor 5\n14 is not divisible by 3 nor 5\n15 is divisible by 3\n15 is divisible by 5\n16 is not divisible by 3 nor 5\n17 is not divisible by 3 nor 5\n18 is divisible by 3\n19 is not divisible by 3 nor 5\n20 is divisible by 5\n\ndivisible by 3 = 5\ndivisible by 5 = 3\ndivisible by neither = 8\n<\/pre><\/div>\n\n\n
\n .. 2 3 .. 5 .. 7 .. .. ..\n 11 .. 13 .. .. .. 17 .. 19 ..\n .. .. 23 .. .. .. .. .. 29 ..\n 31 .. .. .. .. .. 37 .. .. ..\n 41 .. 43 .. .. .. 47 .. .. ..\n .. .. 53 .. .. .. .. .. 59 ..\n 61 .. .. .. .. .. 67 .. .. ..\n 71 .. 73 .. .. .. .. .. 79 ..\n .. .. 83 .. .. .. .. .. 89 ..\n .. .. .. .. .. .. 97 .. .. ..\n<\/pre><\/div>\n\n\n
\n i alpha beta prints\n-----------------------------\n 12 ? ? ?\n<\/pre><\/div>\n\n
\n#include <stdio.h>\n\nint alpha(int a);\nint beta(int b);\n\nint main(void) {\n int i;\n \n for(i = 12; i < 26; i = i + 3)\n if( alpha(i) && beta(i) )\n printf("%d\\n", i);\n}\n\nint alpha(int a) {\n if(a%3 == 0)\n return 1;\n else\n return 0;\n}\n\nint beta(int b) {\n if(b%2 == 0)\n return 1;\n else\n return 0;\n}\n<\/pre><\/div>\n\n\n
\n\n\n\nArrays<\/h3>\n\n\n\n
array1<\/code> in array2<\/code>, but in reverse order. Also write printArray()<\/code> to print the arrays.<\/p>\n\n\n\n#include <stdio.h>\n \nvoid printArray(int [], int);\n\nint main(void) {\n int array1[5] = {17, 23, 9, 4, 12};\n int array2[5];\n int length = 5;\n int i;\n\n \/* code to store array1 in array2 in \n reverse order goes here *\/\n\n printf("array1: ");\n printArray(array1, length);\n printf("array2: ");\n printArray(array2, length);\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n \nvoid printArray(int [], int);\n\nint main(void) {\n int array1[5] = {1, 2, 3, 4, 5};\n int array2[5] = {99, 88, 77, 66, 55};\n int temp;\n int length = 5;\n int i;\n\n \/* code for swapping goes here *\/\n\n printf("array1: ");\n printArray(array1, length);\n printf("array2: ");\n printArray(array2, length);\n}\n\nvoid printArray(int data[], int length) {\n int i;\n\n for(i = 0; i < length; i++)\n printf("%2d ", data[i]);\n\n printf("\\n");\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n \nint main(void){\n int data[2][3] = { {15, 72, 9},\n { 4, 8, 11} };\n int i, k;\n \n for(i = 0; i < 2; i++) {\n for(k = 0; k < 3; k++) {\n printf("%2d ", data[i][k]);\n }\n printf("\\n");\n }\n}\n \n\/*****************\n *** Output ***\n 15 4 \n 72 8 \n 9 11 \n*****************\/\n<\/pre><\/div>\n\n\n\n5, 8, 10\n<\/pre><\/div>\n\n\n
\n5, 15, 10\n<\/pre><\/div>\n\n\n
\n#include <stdio.h>\n\nint main(void)\n{\n int d[] = {9, 8, 2, 15, 6, 32, 10, 4};\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n int d[][3] = {{ 1, 2, 3},\n { 4, 5, 6},\n { 7, 8, 9},\n {-1, -2, -3}};\n int rows;\n\n rows = sizeof(d)\/sizeof(d[0]);\n printRowSums(d, rows);\n} \n<\/pre><\/div>\n\n\n# output\n 1 + 2 + 3 = 6\n 4 + 5 + 6 = 15\n 7 + 8 + 9 = 24\n -1 + -2 + -3 = -6\n<\/pre><\/div>\n\n\n
findNum()<\/code> to search an array for number<\/code>. If it finds number<\/code> in the array, it returns the index of the first occurrence (in case there are multiple occurrences) of number<\/code>. If number<\/code> is not in the array, the function returns -1. Write findNum()<\/code>.<\/p>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n int d[] = {-2, 14, 7, 61, 99, 7, 0, -10};\n int number;\n int index, size = sizeof(d)\/sizeof(d[0]);\n \n number = 7; \/* number to search for *\/\n index = findNum(d, size, number);\n \n if(index >= 0)\n printf("%d is at index %d\\n", number, index);\n \n number = 123; \/* number to search for *\/\n index = findNum(d, size, number);\n \n if(index >= 0)\n printf("%d is at index %d\\n", number, index);\n} \n<\/pre><\/div>\n\n\n
\n\n\n\nVariable Types<\/h3>\n\n\n\n
\n\/*\n on this machine, a short int is 2 bytes, an int \n is 4 bytes, and a long int is 8 bytes\n*\/\n #include <stdio.h>\n \n int main(void) {\n short w = 20000;\n short x = 15000;\n short y;\n short z;\n \n y = w + x;\n z = w - x;\n \n printf("w + x = %hd\\n", y);\n printf("w - x = %hd\\n", z);\n }\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n short int x = 32765;\n int i;\n\n for(i = 0; i < 5; i++) {\n printf("%d\\n", x);\n x++;\n }\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n\nvoid f(int data[]) {\n printf("here data has a size of %d bytes\\n", sizeof(data) );\n}\n\nint main(void) {\n int data[] = {1, 2, 3};\n\n printf("data has a size of %d bytes\\n", sizeof(data) );\n\n f(data);\n}\n\n\/**************************************\n*** Output ***\n data has a size of 12 bytes\n here data has a size of 8 bytes\n*\/\n<\/pre><\/div>\n\n\n
\n\n\n\nPointers<\/h3>\n\n\n\n
\n#include <stdio.h>\n\nint main(void) {\n int x = 5;\n int* x_ptr = &x;\n\n printf("%Z, %Z, %Z, %Z\\n", x, *x_ptr, &x, x_ptr);\n}\n<\/pre><\/div>\n\n\n\nint main(void) {\n int x = 5;\n int *ptrA = &x;\n int **ptrB = &ptrA;\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n \/* 1 2 3 *\/\n \/* 123456789012345678901234567890 *\/\n char fun[] = "Programming is fun.";\n char favorite[] = "I love programming in C.";\n char *x = fun;\n\n printf("%d, ", sizeof(fun));\n printf("%d, ", sizeof(favorite));\n printf("%d\\n", sizeof(x));\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n \nint main(void) {\n int data[] = {1, 2, 3, 4, 5, 6}; \n int i;\n \n for(i = 0; i < 6; i++)\n printf("%2d", data[i]);\n} \n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n int data[] = {1, 2, 3, 4, 5, 6, 7};\n int *ptr = data;\n int i;\n\n printf(" i *ptr\\n--------\\n");\n\n for(i = 2; i > -4; i--) {\n printf("%2d, %2d\\n", i, *ptr);\n ptr += i;\n }\n}\n<\/pre><\/div>\n\n\nvoid<\/code> and the variable y<\/code> gets its value using pointers.<\/p>\n\n\n\n#include <stdio.h>\n\nint dbl(int num);\n\nint main(void) {\n int x = 13;\n\n x = dbl(x);\n printf("x doubled is %d\\n", x);\n}\n\nint dbl(int num) {\n return 2*num;\n}\n<\/pre><\/div>\n\n\ngetRange()<\/code>. getRange()<\/code> is passed the addresses of two variables, low<\/code> and high<\/code>, as well as a variable called mid<\/code>. getRange()<\/code> calculates a value for low<\/code> of 75% of the value of mid<\/code> and a value of 125% of the value of mid<\/code> for the value of high<\/code>. The values of all three variables are printed from main() <\/code>after the function call. Write getRange()<\/code>.<\/p>\n\n\n\n#include <stdio.h>\n\nint main(void) {\n double low, high, mid = 40;\n\n getRange(&low, &high, mid);\n\n printf("low = %5.2f, mid = %5.2f, high = %5.2f\\n", \\ \n low, mid, high);\n}\n\/**********************************************\n*** output ***\n low = 30.00, mid = 40.00, high = 50.00\n*\/\n<\/pre><\/div>\n\n\nprintf()<\/code> statement?<\/p>\n\n\n\n#include <stdio.h>\n \nint main(void) {\n int data[3][4] = { {11, 12, 13, 14},\n {15, 16, 17, 18},\n {19, 20, 21, 22} };\n int i;\n \n for(i = 0; i < 12; i++)\n printf("%d ", *(*data + i) );\n}\n<\/pre><\/div>\n\n\n\n#include <stdio.h>\n \nint main(void) {\n int x = 5;\n int *p1 = &x;\n int **p2 = &p1;\n\n printf("before: x = %d\\n", x);\n change(&p1);\n printf(" after: x = %d\\n", x);\n}\n<\/pre><\/div>\n\n\n
\n\n\n\nStrings<\/h3>\n\n\n\n