Worksheets are self-guided activities that reinforce lectures. They are not graded for accuracy, only for completion. Worksheets are due by Sunday night before the next lecture.
-
Examine the program below and answer the following questions.
#include <stdio.h>
int main() {
char a[3] = "Go Buff!";
printf("%s\n", a);
}
Attempt to answer these questions without running the code!
- Explain why the full string
"Go Buff!" will not print.
- Provide two ways to fix the program so the full string would print.
Reveal Solution
- Because
a is declared to have length 3, it only stores the first 3 characters of "Go Buff!" which are "Go ".
- Method 1: Change the declaration of
a to char a[9] = "Go Buff!"; because there are 8 characters in "Go Buff!".
- Method 2: Change the declaration of
a to char a[] = "Go Buff!"; so we do not explicitly declare the length to the string a.
- Method 3: Change the declaration of
a to char *a = "Go Buff!"; so we have a pointer to a string.
-
Examine the program below and answer the following questions.
#include <stdio.h>
#include <string.h>
int main() {
char str[] = "string";
if (sizeof(str) == strlen(str)) {
printf("equal\n");
} else if (sizeof(str) > strlen(str)) {
printf("greater than\n");
}
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- How is the sizeof function different from the strlen function?
Reveal Solution
- The output of this program is
greater than.
- The sizeof function gives how much memory is used (in terms of bytes) to store
str, which is 7 because there is a null terminator at the end of "string".
- The strlen function simply gives the length of the string
str, which is 6 because there are 6 characters in "string".
-
Consider the following code snippet
#include <stdio.h>
int main(){
char str[] = "A string";
printf("sizeof(str) = %ul\n",sizeof(str));
printf("strlen(str) = %d\n",strlen(str));
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- Why does the sizes differ?
Reveal Solution
- The output of this program is
sizeof(str) = 9
strlen(str) = 8
- The sizes are different because the
sizeof function includs the terminating symbol (‘\0’) of str while the strlen function returns the number of characters in the string ignoring the terminating symbol. Consequently, sizeof(str) returns 9 while strlen(str) returns 8 since there are 8 symbols in string "A string".
-
Consider the following code snippet
#include <stdio.h>
int main(){
char * str = "A string";
printf("sizeof(str) = %ul\n",sizeof(str));
printf("strlen(str) = %d\n",strlen(str));
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- Why does the sizes differ?
Reveal Solution
- The output of this program is
sizeof(str) = 8
strlen(str) = 8
- The sizes happens to be the same.
sizeof(str) is 8 due to the actual size of pointers and strlen(str) is 8 due to there are 8 characters in str.
-
Examine the program below and answer the following questions.
#include <stdio.h>
#include <string.h>
int main() {
char a[] = "aaa";
char b[] = "bbb";
if (strcmp(a, b) == -1) {
printf("aaa is less than bbb\n");
}
if (strcmp(b, a) == 1) {
printf("bbb is greater than aaa\n");
}
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- Explain how function strcmp from the string.h library works.
Reveal Solution
- The output of this program is
aaa is less than bbb
bbb is greater than aaa
- The function strcmp takes 2 string arguments, s1 and s2 e.g.
strcmp(s1, s2). The return value is 1 when s1 is greater alphabetically, is 0 when the two strings are identical and is -1 when s1 is smaller alphabetically.
-
Examine the program below and answer the following questions.
#include <stdio.h>
#include <string.h>
int main() {
char str1[] = "10 9 8 7 6 5 4 3 2 1";
char * str2 = "10 9 8 7 6 5 4 3 2 1";
if (strcmp(str1, str2) == 0) { //tests if they are the same string
printf("Same\n");
}
else{
printf("Different\n");
}
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- Describe how a char * and an char array are similar.
Reveal Solution
- The output of this program is
Same.
- Both char * and char array can be used to represent a string.
-
Examine the program below and answer the following questions.
#include <stdio.h>
#include <string.h>
int main() {
char str1[10] = "10 9 8 7 6 5 4 3 2 1";
char * str2 = "10 9 8 7 6 5 4 3 2 1";
if (strcmp(str1, str2) == 0) { //tests if they are the same string
printf("Same\n");
}
else{
printf("Different\n");
}
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- Why is it different then the output from the previous question.
Reveal Solution
- The output of this program is
Different.
- The output is different from the previous question because
str1 was declared to have length 10 which means it only contains the string 10 9 8 7 6. This is different from str2, which equal to 10 9 8 7 6 5 4 3 2 1.
-
The following program is WRONG!
#include <stdio.h>
#include <string.h>
int main() {
char str[100] = "C is my favorite language";
str[strlen(str)+0] = '!';
str[strlen(str)+1] = '!';
str[strlen(str)+3] = '\0';
printf("%s And python is my second!\n", str);
}
We want the output to be
C is my favorite language!! And python is my second!
But the output is this instead (note the single !)
C is my favorite language! And python is my second!
Answer the following questions:
- Why did the program not function as expected?
- Provide a corrected version of the program
Reveal Solution
strlen(str) is incrementing as we add characters to str. The first time we do str[strlen(str)+0], character ! is added to the end of the string. Consequently, strlen(str) also incremented by 1, from 25 to 26. Then, we want to add another ! by doing str[strlen(str)+1] which did not work because at this point, strlen(str)+1 is 26 and adding to the 27th character will result in having a character after the terminating character.- Fixes for the program can be
#include <stdio.h>
#include <string.h>
int main() {
char str[100] = "C is my favorite language";
int len = strlen(str); // New line fixing the code
str[len+0] = '!'; // Changing strlen(str) to the new variable len
str[len+1] = '!'; // Changing strlen(str) to the new variable len
str[len+3] = '\0'; // Changing strlen(str) to the new variable len
printf("%s And python is my second!\n", str);
}
or
#include <stdio.h>
#include <string.h>
int main() {
char str[100] = "C is my favorite language";
str[strlen(str)] = '!'; // Changing strlen(str)+0 to strlen(str)
str[strlen(str)] = '!'; // Changing strlen(str)+1 to strlen(str)
str[strlen(str)] = '\0'; // Changing strlen(str)+2 to strlen(str)
printf("%s And python is my second!\n", str);
}
-
Examine the program below and answer the following questions.
#include <stdio.h>
#include <string.h>
int main() {
char * strings[3] = {"Hello", " world", " !!!"};
char o = 'o';
char * o_str = "o";
char * world = "world";
printf("P1: ");
if (strings[0][4] == o) {
printf("true\n");
} else {
printf("false\n");
}
printf("P2: ");
if (*strings[0] + 4 == *o_str) {
printf("true\n");
} else {
printf("false\n");
}
printf("P3: ");
if (strcmp((strings[1] + 1), world) == 0) {
printf("true\n");
} else {
printf("false\n");
}
printf("P4: ");
if (strings[2] == " !!!") {
printf("true\n");
} else {
printf("false\n");
}
}
Attempt to answer these questions without running the code!
- What is the output for P1, P2, P3, and P4?
- Please explain what exactly each if statement is checking, as in what is being compared and why they are equal or not.
Reveal Solution
- The output of this program is
P1: true
P2: false
P3: true
P4: true
P1: true It is checking if the 5th character of the 1st string (array indexes starts from zero) in strings is the character “o”. The result is true because the character is indeed “o” and == can be used for character comparison as it compares the ASCII code of the characters on each side.P2: false Firstly, *strings[0] dereferences the first character in the first string of strings (“Hello”), which is the character “H”, adding 4 to it will result in ‘L’ (4th character after ‘H’ in the ASCII Table). The right hand side is dereferencing a pointer to character “o”, which is the character “o”. Thus, the left and right hand sides are not equal.P3: true We first observe that strings[1] is the string “ world”, with a space as the first character and an implicit “\0” terminating the string. strings[1]+1 returns the string “world” because it starts at the 2nd character of strings[1] and terminates where strings[1] terminates. Thus, we are comparing “world” with “world” using the strcmp method and the result is the two sides are equal, thus true.P4: true This is tricky! When you declare a string using char *, you are creating a constant array in the data segment. So if there are two strings in the program declared in this way, then C optimizes and has each case reference the same instance of the string in the same memory location. In this case strings[2] references the string " !!!" which is compared to a constant string " !!!" in the if statement. C looks at this and notes that these two strings can be the same in memory, and does so. Thus when we compare the pointers, they are the same pointer value because they point to the same string. The result is true.
-
Examine the program below and answer the following questions.
#include <stdio.h>
int main() {
char str[100] = "the importance of being earnest";
str[0] -= 32;
str[4] -= 32;
str[18] -= 32;
str[24] -= 32;
printf("%s\n", str);
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- How did this happen?
Reveal Solution
- The output of this program is
The Importance of Being Earnest.
- In the ASCII table, code for lower case letters are larger than their corresponding upper case by 32. e.g. a’s ASCII code is 97 whereas A’s is 65. Thus, in the program, we took the four letters from 4 different positions (0, 4, 18 and 24) and turning them into the same letter but in upper case.
-
Examine the program below and answer the following questions.
#include <stdio.h>
void super_cool_function(char *p) {
*p = *p - 4;
}
int main() {
char a[6] = {'L','I','P','P','S', 4};
for (int i = 0; i < 6; i++) {
char * b = &a[i];
super_cool_function(b);
}
printf("%s\n", a);
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- What does super_cool_function do?
Reveal Solution
- The output of this program is
HELLO.
- The super_cool_function takes a pointer to charater as a single parameter
char *p. It dereference the pointer *p, decrease the value by 4 *p - 4 and put the value into the slot where *p is pointing to. Note that decreasing a character’s value by 4 means going back 4 characters in alphbetical order (ASCII Table).
-
Examine the program below and answer the following questions.
#include <stdio.h>
#include <string.h>
int main() {
char string_a[24] = "Raise High";
char * string_b = " the Roof Beam, Carpenters";
printf("%s\n", strncat(string_a, string_b, 14));
}
Attempt to answer these questions without running the code!
- What is the output of this program?
- What is the function definition of the C string.h function strncat. (Hint: use the man pages!)
Reveal Solution
- The output of this program is
Raise High the Roof Beam.
- strncat appends the second string (source string - first parameter) to the first string (destination string - second parameter) with a maximum of n (third parameter) bytes from the source string. In this case, 14 bytes from
string_b is appended into the destination string string_a. Note that string_a has 24 characters/bytes of space, which perfectly fits the resulting string of Raise High the Roof Beam from the strncat function. If the destination string is not large enough for the result, an error will occur.
-
Consider the following code snippet
#include <stdio.h>
#include <string.h>
int main(){
char str[] = "this is a string";
for(char * p = str; *p ; p++){//<-- A
if(*p > 'l'){
*p = *p+1; //<-- B
}
}
}
Attempt to answer these questions without running the code!
- For the code line marked as
A, explain why the for loop will properly iterate over each character in the string str.
- For the code line marked as
B, explain how this will modify the string
- What is the expected output?
Reveal Solution
- The for loop is turned into the following while loop
char * p = str;
while(*p) {
// code
p++;
}
Where character pointer p is initialized with the first address of str. Until p becomes the terminating character at the end of str, the while loop (for loop) will keep running and within each loop, p is incremented by 1, which makes it point to the next address (stores the next character of str).
- Line marked as
B will make the character in the string the next character alphabetically (e.g. a->b, y->z). It first dereference p, add 1 to the value and return it to the slot p is pointing to.
uhit it a tusiog
-
Consider the following code snippet
int mystery(char * s1, char * s2){
while( *s1++ == *s2++ && *s1 && *s2);//<-- A
return( *s1 == '\0' && *s2 == '\0' );//<-- B
}
Attempt to answer these questions without running the code!
- What does the
mystery function do?
- Offer an explanation for the functionality of line
A
- Offer an explanation for the functionality of line
B
Reveal Solution
- The
mystery function compares two strings s1 and s2, returns 1 if they are identical and 0 if not.
- Line
A loops through the two strings, until at least 1 of the following 3 things happen:
- The character
s1 is pointing to does not equal to the character s2 is pointing to.
- The character
s1 is pointing to is the terminating character “\0”.
- The character
s2 is pointing to is the terminating character “\0”.
- Line
B returns 1 (true) only when both s1 and s2 are pointing to the terminating character “\0”, otherwise it returns 0 (false).
- Note: The function above will not work if
s1 and s2 varies only by the last character. The code below will work as intended by comparing the last character in the ruturn statement (line B).
int mystery(char * s1, char * s2){
while( *s1++ == *s2++ && *s1 && *s2);//<-- A
return( *(s1-1)==*(s2-1) && *s1 == '\0' && *s2 == '\0' );//<-- B
}
-
Examine the program below and complete the code.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void print_error_and_exit(void) {
fprintf(stderr, "USAGE ERROR - try again like this:\n");
fprintf(stderr, "count -to <number> -by <number>\n");
exit(1); //exit the program
}
int main(int argc, char * argv[]) {
int to_num;
int by_num;
// Validate argument count
if (argc < 5 || argc > 5) {
print_error_and_exit();
}
// Validate -to parameter
if (strcmp(argv[1], "-to") != 0) {
print_error_and_exit();
}
// Validate -to parameter number
if (sscanf(argv[2], "%d", &to_num) == 0) {
print_error_and_exit();
}
// COMPLETE THE PARAMETER VALIDATION CODE
for (int count = by_num; count <= to_num; count += by_num) {
printf("%d\n", count);
}
}
Finish the code above so that the output is the following:
$ ./count bad parameters
USAGE ERROR - try again like this:
count -to <number> -by <number>
$ ./count -to 10 -by 2
2
4
6
8
10
Reveal Solution
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void print_error_and_exit(void) {
fprintf(stderr, "USAGE ERROR - try again like this:\n");
fprintf(stderr, "count -to <number> -by <number>\n");
exit(1); //exit the program
}
int main(int argc, char * argv[]) {
int to_num;
int by_num;
// Validate argument count
if (argc < 5 || argc > 5) {
print_error_and_exit();
}
// Validate -to parameter
if (strcmp(argv[1], "-to") != 0) {
print_error_and_exit();
}
// Validate -to parameter number
if (sscanf(argv[2], "%d", &to_num) == 0) {
print_error_and_exit();
}
// Start of Solution Code
// Validate -by parameter
if (strcmp(argv[3], "-by") != 0) {
print_error_and_exit();
}
// Validate -by parameter number
if (sscanf(argv[4], "%d", &by_num) == 0) {
print_error_and_exit();
}
// End of Solution Code
for (int count = by_num; count <= to_num; count += by_num) {
printf("%d\n", count);
}
}