Check chars in a string while using pointers

Question

I want to check each individual char in a string. The string is stored in a pointer. for some reason I can't, it only let me get the whole string. here's my code:

int main() {
  char *s=(char*)calloc(30,sizeof(char));
  s="hello";
  printf("%s",&s[2]);
  return 0;
 }

this code print "llo", i need only 1 char like "l" or "o". anyone know how i can achieve it? ty

Printf characters not strings. Look up the formatting codes for printf, and ask yourself what is the value of s[2] if you leave off the '&'. — Dave S
– Dave S, Commented Jun 14, 2020 at 17:40
you also lost the block you created with calloc (memory leak). What do you mean by check each individual char ? — bruno
– bruno, Commented Jun 14, 2020 at 17:41
try doing strcpy instead, setting s to point to "hello" is not the same thing as copying the string. — AndersK
– AndersK, Commented Jun 14, 2020 at 19:21
check each individual char, i mean like if i want to find the letter "l" or check if all the string is numbers only or none numbers string. — operout
– operout, Commented Jun 15, 2020 at 10:06

RobertS supports Monica Cellio · Accepted Answer · 2020-06-14 19:00:56Z

2

Use the %c conversion specifier to print a sinlge character instead of %s to print a string.

Also the memory allocation by calloc() is useless, since the pointer to char s get assigned by the address of the first element of the string literal "hello" one statement later.

#include <stdio.h>

int main (void) 
{
    const char *s = "hello";
    printf("%c", s[2]);
    return 0;
}

Output:

Side Note:

Use the const qualifier to prevent any unintentional write attempts to the string literal that lead to undefined behavior.

If you want to allocate memory and assign/initialize the allocated memory by the string literal "hello", use strcpy() (header string.h):

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

int main (void) 
{
    char *s = calloc(30, sizeof(*s));
    if (!s)      // Check if allocation failed.
    {
        fputs("Error at allocating memory!", stderr);
        exit(1);
    }
    strcpy(s, "hello");
    printf("%c", s[2]);
    return 0;
}

Output:

or you can use strdup() (Note: strdup() is not part of the standard C library) which will automatically allocate memory with the string of the string literal passed as argument initialized:

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

int main (void) 
{
    char *s = strdup("hello");
    if (!s)            // Check if allocation failed.
    {
        fputs("Error at allocating memory!", stderr);
        exit(1);
    }

    printf("%c", s[2]);
    return 0;
}

Side note:

"The string is stored in a pointer."

Something like that is not possible. The pointer points to the first element of the string (literal). The pointer does not store a string.

edited Jun 14, 2020 at 19:00

answered Jun 14, 2020 at 17:47

RobertS supports Monica Cellio

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4 Comments

pm100 Over a year ago

or even strdup("Hello")

RobertS supports Monica Cellio Over a year ago

@pm100 Problem is that strdup() isn't part of the standard. I'm also not sure whether OP only allocates the memory for the string literal "hello". Then s/he would not allocate an buffer of 30 char elements and only of 6 char's. Quite mysterious code, tough.

pm100 Over a year ago

strdup is posix standard

RobertS supports Monica Cellio Over a year ago

@pm100 Took it into the answer. Thanks. Was also a little absent before. OP can of course realloc() the malloc if needed.

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