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How to iterate over a string in C?

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c

iteration

People also ask

Can we iterate a string?

Another way to iterate over a string is to use for item of str . The variable item receives the character directly so you do not have to use the index. If your code does not need the index value of each character, this loop format is even simpler.

Can I iterate through string C++?

Program to loop on every character in string in C++To loop on each character, we can use loops starting from 0 to (string length – 1). For accessing the character we can either use subscript operator "[ ]" or at() function of string object.

How do you print a loop in a string?

C Program to Print Characters in a String Example 1 This program allows the user to enter a string (or character array). Next, we used While Loop to iterate each character inside a string. Inside this, we used the printf statement to print characters in this string. The condition is True because str[0] = h.


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++) {

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.

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;
}

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().


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.


  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.

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]);

    }


 }