如何在C中将一个char链接为一个char *？

Question

How can I prepend char c to char* myChar ? I have c has a value of "A" , and myChar has a value of "LL" . How can I prepend c to myChar to make "ALL" ?

Answer 1

This should work:

#include <string.h>    
char * prepend(const char * str, char c)
{
    char * new_string = malloc(strlen(str)+2);  // add 2 to make room for the character we will prepend and the null termination character at the end
    new_string[0] = c;
    strcpy(new_string + 1, str);
    return new_string;
}

Just remember to free() the resulting new string when you are done using it, or else you will have a memory leak.

Also, if you are going to be doing this hundreds or thousands of times to the same string, then there are other ways you can do it that will be much more efficient.

Answer 2

Well, first you need to figure out storage for your new string. If all I know about the source is a char * , I don't know whether it points to a buffer that I can overwrite and/or is long enough, or whether I need to allocate a new buffer.

To allocate a new buffer, it'd be something like this:

char * strprep(char c, const char * myChar)
{
    /* length, +1 for c, +1 for '\0'-terminator */
    char * newString = malloc(strlen(myChar) + 2);

    /* out-of-memory condition rolls downhill */
    if (newString == NULL) return NULL;

    /* populate the new string */
    newString[0] = c;
    strcpy(newString + 1, myChar);
    return newString;
}

To overwrite in place, you'll need to move the characters over to make room for the new start of string:

char * strprep(char c, char * myChar)
{
    int len = strlen(myChar);
    int i;

    /* Move over! */
    for (i = len; i >= 0; i--) myChar[i + 1] = myChar[i];

    /* Now plug in the new prefix. */
    myChar[0] = c;
    return c;
}

If your C library has it available, you can use memmove in place of the loop to shift the original characters up by one.

Answer 3

Probably not the most efficient but this is a step in the right direction: this one is clearly direct approach -

  int main()
   {
      int i;
      char tempc;
      for (i=0; i<strlen(str); i++)
      {
        tempc = c;
        c = str[i];
        str[i] = tempc;            
         str[i+1] = c;
      }
       *(str[i+1]) = malloc(1);
       str[i+1] = '\0';
    }