How to iterate over a string in C?

Question

Right now I\'m trying this:

#include 

int main(int argc, char *argv[]) {

    if (argc != 3) {

        printf(\"Usage: %s %s sourcecode inpu         


        

Answer 1

Just change sizeof with strlen.

Like this:

char *source = "This is an example.";
int i;

for (i = 0; i < strlen(source); i++){

    printf("%c", source[i]);

}

Answer 2

One common idiom is:

char* c = source;
while (*c) putchar(*c++);

A few notes:

• In C, strings are null-terminated. You iterate while the read character is not the null character.
• *c++ increments c and returns the dereferenced old value of c.
• printf("%s") prints a null-terminated string, not a char. This is the cause of your access violation.

Answer 3

This should work

 #include <stdio.h>
 #include <string.h>

 int main(int argc, char *argv[]){

    char *source = "This is an example.";
    int length = (int)strlen(source); //sizeof(source)=sizeof(char *) = 4 on a 32 bit implementation
    for (int i = 0; i < length; i++) 
    {

       printf("%c", source[i]);

    }


 }

Answer 4

sizeof(source) returns sizeof a pointer as source is declared as char *. Correct way to use it is strlen(source).

Next:

printf("%s",source[i]); 

expects string. i.e %s expects string but you are iterating in a loop to print each character. Hence use %c.

However your way of accessing(iterating) a string using the index i is correct and hence there are no other issues in it.

Answer 5

Replace sizeof with strlen and it should work.

Answer 6

You need a pointer to the first char to have an ANSI string.

printf("%s", source + i);

will do the job

Plus, of course you should have meant strlen(source), not sizeof(source).