I am trying to make this: if I input a string, it wil print the same. But if I type quit as input, the program will stop printing.
#include <stdio.h>
int main() {
char input[11];
printf("Give your input: \n");
scanf("%s", &input);
while (input!= "quit") {
printf("%s\n", input);
scanf("%s", &input);
}
return 0;
}
There is nothing wrong syntactically in your code (i.e. it compiles cleanly), but logically it will never do what you expect it to do. The == and != operators in C are good for checking that two logical values are equivalent or not. Eg. say the user enters the value "quit" into stdin. then the expression input != "quit" means that != will be evaluate the addresses of each buffer. for illustration, it would be equivalent to the following:
unsigned int address1 = &input[0];//uaddress of input char array
unsigned int address2 = &"quit"[0];//address of string literal
while(address1 != address2) {
The two strings will never be located in the same location, and the while(){ will never exit.
The proper way to compare strings in C is by using one of the string compare functions, strstr() or strcpy().
But then this brings us to the second part of the problem. A string buffer populated using scanf() with a format specifier: "%s" will always include the last character the user typed in, i.e. the newline character: \n. So what you believed to be "quit" is actually "quit\n". Resulting in a failed string compare. (if using strcmp(), which in turn will cause your loop to never exit:
If input contains "quit\n" and the string literal (2nd argument) contains quit, the expression: (Again, the user enters "quit", it is captured into input as "quit\n")
while(strcmp(input, "quit") != 0 { //will not exit
The core issue here: Change:
while (input!= "quit") {
To:
while (strcmp(input,"quit") != 0) {
The following code addresses this and other issues below. The following example uses strcmp().
Modified code with comments:
#include <stdio.h>
int main() {
char input[50];//use a bigger value for array
printf("Give your input: \n");
scanf("%49s", input);//scanf picks up the newline
// | added width specifier to prevent overflow, and removed &
input[strcspn(input, "\n")] = 0;//removes newline character
while (strcmp(input,"quit") != 0) {//use string compare function"
printf("%s\n", input);
printf("Give your input: \n");//prompt user each interation
scanf("%49s", input);//remove '&'
// | added width specifier to prevent overflow, and removed &
input[strcspn(input, "\n")] = 0;//removes newline character
}
return 0;
}
The compiler can tell you about mistakes you've made. The format strings are warnings by default, but the mistake of comparing strings is only reported (in the version of gcc I'm using) if you use the flag -Wall. I suggest you always use it, and always fix problems that cause warnings.
$ gcc -Wall debug.c
debug.c: In function ‘main’:
debug.c:7:13: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[11]’ [-Wformat=]
7 | scanf("%s", &input);
| ~^ ~~~~~~
| | |
| | char (*)[11]
| char *
debug.c:9:17: warning: comparison with string literal results in unspecified behavior [-Waddress]
9 | while (input!= "quit") {
| ^~
debug.c:11:17: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[11]’ [-Wformat=]
11 | scanf("%s", &input);
| ~^ ~~~~~~
| | |
| | char (*)[11]
| char *
The compiler doesn't tell you to use strcmp (perhaps it should), but the warnings flag a problem. And it's a much better stack-overflow question to say "the compiler gave me this specific warning on this specific code, but I don't understand it. Can someone explain?"
