反转C字符串函数会崩溃吗？

Question

I'm trying to write a C function to reverse a passed in C style string (ie char *) and return the char pointer of the reversed string. But when I run this in VS2012, nothing is printed in terminal and "main.exe has stopped working" msg shows up.

#include <stdio.h>
#include <string.h>
char * rrev_str(char * str )
{
    char *revd_str=""; //I tried char revd_str []="" error: stack around "revd_str" is corrupted
    int i,r;
    int str_len=strlen(str);
    for (i = str_len-1, r=0; i >=0; i--,r++)
    {
        revd_str[r]= str[i];
    }
    return revd_str;
}

int main(int argc, char* argv[])
{
   char str1 [] ="STEETS";
   char str2 [] ="smile everyday!";

   //reverse "chars" in a C string and return it
   char * rev_string=rrev_str(str1);
}

Answer 1

The problem here is three fold. First you aren't allocating enough space for the reversed string, and secondly you are returning a pointer to a local variable in rrev_str(), and thirdly you're modifying a string literal. You need to allocate space for revd_str on the heap:

char * rrev_str(char * str )
{   
    int i,r;
    int str_len=strlen(str);

    char *revd_str=malloc(str_len + 1); 
    memset(revd_str, 0, str_len + 1);

    for (i = str_len-1, r=0; i >=0; i--,r++)
    {
        revd_str[r]= str[i];
    }
    return revd_str;
}

Answer 2

Problem:

You are accessing invalid memory address.
revd_str is pointing to literal constant string of length 1 and you are accessing it beyond the length which is invalid.

Solution:

• Create char array of require length (statically or dynamically).
• Reverse the given string.
• Pass 2nd param as destination string
  syntax: char * rrev_str(char * src, char *dest);

Reverse the given string

char * rrev_str(char * str )
{
   int start = 0;
   int end = strlen(str) - 1;
   char temp;

    for (; start < end; start++ ,end--)
    {
        temp = str[start];
        str[start] = str[end];
        str[end] = temp;
    }
    return str;
}

int main(int argc, char* argv[])
{
   char string [] ="smile";

   //reverse "chars" in a C string and return it
   char * rev_string = rrev_str(string);

   printf("%s",rev_string);
}

Pass 2nd param as destination string

char * rrev_str(char * src, char *dest)
{
   int srcLength = strlen(src);
   int destLength = strlen(dest);
   int i;
   // Invalid destination string
   if (srcLength > destLength)
   {
        return NULL;
   }

   dest[srcLength] = '\0';
   srcLength--;
    for (i=0; srcLength >= 0;i++, srcLength--)
    {
        dest[i] = src[srcLength];
    }

 return dest;
}

int main(int argc, char* argv[])
{
   char string [] ="smile";
   char revString[20];  

   //reverse "chars" in a C string and return it
   char * rev_string = rrev_str(string, revString);

    printf("%s",rev_string);
}

Answer 3

What! you are doing..

char *revd_str=""; // Creating String Literal which can't be modified because they are read only  
char *revd_str[]=""; // Creating Char Array of Size Zero.

So Solution are

Either take reference of your string

char *revd_str = strdup(str);

Or create dynamic char array

char *revd_str = (char*) malloc (strlen(str)+1);  

your program will run fine. logic is incorrect for reversing so modify it. A sample solution is given below

char * rrev_str(char * str )
{
    char *revd_str=strdup(str);
    int i;  // no need for extra 'int r'
    int str_len=strlen(str);
    for (i = 0; i < str_len/2; i++)
    {
        char temp = revd_str[i];
        revd_str[i]= revd_str[str_len - 1 -i];
        revd_str[str_len - 1 -i] = temp;
    }
    return revd_str;
}