如何使用strcpy（）函数将字符存储到char指针中

Question

I am trying to figure out why I can't store characters into my char pointer by using the strcpy() command. I get a seg fault when I run the code below.

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

int main(int argc, const char *argv[])                                               
{                                                                                    
   char *str1, *str2;                                                               
   int ret;                                                                         

   strcpy(str1, "abcdefg"); // stores string into character array str1              
   strcpy(str2, "abcdefg");                                                         

   printf("contents of %s \n", str1);                                               

   ret = strncmp(str1, str2, strlen(str2)); /* compares str1 to str2 */             

   if (ret > 0) {                                                                   
       printf("str1 is less than str2\n");                                          
  }                                                                                
  else if (ret < 0) {                                                              
      printf("str2 is less than str1\n");                                          
  }                                                                                
  else if (ret == 0) {                                                             
      printf("str1 is equal to str2\n");                                           
  }                                                                                

  return 0;                                                                        
  }   

Thank you!

Answer 1

Right now, str1 and str2 are just pointers to a character. When you do strcpy(str1, "abcdefg") , it attempts to write the characters in the string "abcdefg" into the memory that str1 points to and since str1 points to an unknown memory, which probably you don't have any write permissions, you get a segmentation fault.

One way to fix it is to allocate memory on the heap and then store these strings.

#include <stdlib.h>
...
/* assuming the max length of a string is not more than 253 characters */
char *str1 = malloc(sizeof(char) * 254);
char *str2 = malloc(sizeof(char) * 254);

You can also use strdup to duplicate the string like Gangadhar has mentioned.

Another way is to declare str1 and str2 as arrays during the compiling as Bryan Ash suggested

char str1[] = "abcdefg";              
char str2[] = "abcdefg";

In case, you want to allocate the string dynamically but not on the heap, you can use alloca (for more details read http://man7.org/linux/man-pages/man3/alloca.3.html ) as kfsone noted

Answer 2

compiling this with the -Wall command gives a useful hint

test.c:12:10: warning: 'str1' is used uninitialized in this function [-Wuninitialized]
    printf("contents of %s \n", str1);
          ^
test.c:14:17: warning: 'str2' is used uninitialized in this function [-Wuninitialized]
    ret = strncmp(str1, str2, strlen(str2)); /* compares str1 to str2 */

Answer 3

Given this example, you don't even need strcpy, you could use:

char str1[] = "abcdefg";              
char str2[] = "abcdefg";

If you want to learn more about pointers, there's an excellent free e-book titled Pointers and Memory from Stanford CS Education Library.