如何在字符串中间找到一个字符？

Question

I want to find 3 characters in a string.

I already use strchr and strrchr but I cannot get the middle char.

For example:

hehoho

If I use strchr I get the first h .

If I use strrchr I get the last h .

How can I get the the middle h : he oho ?

char *p;
char* q;
int loc2;
int loc;
char ch[100];
char s[100] = "heheho"

p = strrchr(s, ch);
q = strchr(s, ch);

loc = (int)(p - s);
loc2 = (int)(q - s);

if ( q != NULL)
{   
    a[loc] = ch;
    a[loc2] = ch;
}

Answer 1

One way is to keep calling strchr until you run out of found characters. The function returns NULL when the character isn't found, so maybe something like this:

for(const char* ptr=s; ptr!=NULL && *ptr!='\0'; ptr=strchr(ptr+1, key))
{
  index[found_count] = ptr-s;
  found_count++;
}

This stores down the array index of every found character that matches the character "key", then keeps searching until it finds no more matches and strchr returns NULL. So it finds all occurrences.

Each time you tell strchr to look at the current position + 1. If not for the + 1, you'd keep getting the same result in infinity.

The check *ptr!='\0' for null termination isn't really necessary but catches the case of an empty input string.

ptr-s is pointer arithmetic and gives a integer type corresponding to the position in the string. It will only get executed when the pointer actually points at a valid array item.

Full example:

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

int main (void)
{
  const char s[100] = "heheho";
  const char key = 'h';
  int found_count = 0;
  int index[10];

  for(const char* ptr=s; ptr!=NULL && *ptr!='\0'; ptr=strchr(ptr+1, key))
  {
    index[found_count] = ptr-s;
    found_count++;
  }

  for(size_t i=0; i<found_count; i++)
  {
    printf("Found %c at index %d.\n", key, index[i]);
  }

  printf("So the middle one is at index %d.", index[found_count/2]);
}

Output:

Found h at index 0.
Found h at index 2.
Found h at index 4.
So the middle one is at index 2.