How to iterate over a string in C?

57

Right now I'm trying this:

#include <stdio.h>

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

    if (argc != 3) {

        printf("Usage: %s %s sourcecode input", argv[0], argv[1]);
    }
    else {
        char source[] = "This is an example.";
        int i;

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

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

    getchar();

    return 0;
}

This does also NOT work:

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

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

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

I get the error

Unhandled exception at 0x5bf714cf (msvcr100d.dll) in Test.exe: 0xC0000005: Access violation while reading at position 0x00000054.

(loosely translated from german)

So what's wrong with my code?

c
iteration
asked on Stack Overflow Jul 9, 2010 by Vincent • edited May 26, 2015 by Peter Mortensen

13 Answers

61

You want:

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

sizeof gives you the size of the pointer, not the string. However, it would have worked if you had declared the pointer as an array:

char source[] = "This is an example.";

but if you pass the array to function, that too will decay to a pointer. For strings it's best to always use strlen. And note what others have said about changing printf to use %c. And also, taking mmyers comments on efficiency into account, it would be better to move the call to strlen out of the loop:

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

or rewrite the loop:

for (i = 0; source[i] != 0; i++){
answered on Stack Overflow Jul 9, 2010 by (unknown user) • edited Jul 9, 2010 by (unknown user)
46

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.
answered on Stack Overflow Jul 9, 2010 by Alexandre C. • edited Sep 3, 2017 by Dave Yarwood
3

Rather than use strlen as suggested above, you can just check for the NULL character:

#include <stdio.h>

int main(int argc, char *argv[])
{
    const char *const pszSource = "This is an example.";
    const char *pszChar = pszSource;

    while (pszChar != NULL && *pszChar != '\0')
    {
        printf("%s", *pszChar);
        ++pszChar;
    }

    getchar();

    return 0;
}
answered on Stack Overflow Jul 9, 2010 by Mark Ingram • edited Apr 3, 2013 by Mark Ingram
2

sizeof(source) returns the number of bytes required by the pointer char*. You should replace it with strlen(source) which will be the length of the string you're trying to display.

Also, you should probably replace printf("%s",source[i]) with printf("%c",source[i]) since you're displaying a character.

answered on Stack Overflow Jul 9, 2010 by Jacob
2

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]);

    }


 }
answered on Stack Overflow Jul 9, 2010 by Prasoon Saurav • edited Jul 9, 2010 by Prasoon Saurav
2
  1. sizeof() includes the terminating null character. You should use strlen() (but put the call outside the loop and save it in a variable), but that's probably not what's causing the exception.
  2. you should use "%c", not "%s" in printf - you are printing a character, not a string.
answered on Stack Overflow Jul 9, 2010 by KenE
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]);

}
answered on Stack Overflow Jul 9, 2010 by Pablo Santa Cruz • edited Jul 9, 2010 by Pablo Santa Cruz
1
  • sizeof(source) is returning to you the size of a char*, not the length of the string. You should be using strlen(source), and you should move that out of the loop, or else you'll be recalculating the size of the string every loop.
  • By printing with the %s format modifier, printf is looking for a char*, but you're actually passing a char. You should use the %c modifier.
answered on Stack Overflow Jul 9, 2010 by JSBձոգչ
0

Replace sizeof with strlen and it should work.

answered on Stack Overflow Jul 9, 2010 by Keith Randall
0

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).

answered on Stack Overflow Jul 9, 2010 by ULysses • edited May 26, 2015 by Peter Mortensen
0

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.

answered on Stack Overflow Jul 9, 2010 by Praveen S
0

The last index of a C-String is always the integer value 0, hence the phrase "null terminated string". Since integer 0 is the same as the Boolean value false in C, you can use that to make a simple while clause for your for loop. When it hits the last index, it will find a zero and equate that to false, ending the for loop.

for(int i = 0; string[i]; i++) { printf("Char at position %d is %c\n", i, string[i]); }
answered on Stack Overflow Jun 22, 2019 by Edwin Jata • edited Jun 26, 2019 by Edwin Jata
0

An optimized approach:

for (char character = *string; character != '\0'; character = *++string)
{
    putchar(character); // Do something with character.
}

Most C strings are null-terminated, meaning that as soon as the character becomes a '\0' the loop should stop. The *++string is moving the pointer one byte, then dereferencing it, and the loop repeats.

The reason why this is more efficient than strlen() is because strlen already loops through the string to find the length, so you would effectively be looping twice (one more time than needed) with strlen().

answered on Stack Overflow Dec 28, 2019 by Juniorized

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