Concatenate two char* strings in a C program

Question

I wrote the following C program:

int main(int argc, char** argv) {

    char* str1;
    char* str2;
    str1 = \"sssss\";
    str2 = \"kkkk\";
    printf(\"%         


        

Answer 1

When you use string literals, such as "this is a string" and in your case "sssss" and "kkkk", the compiler puts them in read-only memory. However, strcat attempts to write the second argument after the first. You can solve this problem by making a sufficiently sized destination buffer and write to that.

char destination[10]; // 5 times s, 4 times k, one zero-terminator
char* str1;
char* str2;
str1 = "sssss";
str2 = "kkkk";
strcpy(destination, str1);
printf("%s",strcat(destination,str2));

Note that in recent compilers, you usually get a warning for casting string literals to non-const character pointers.

Answer 2

Here is a working solution:

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

int main(int argc, char** argv) 
{
      char str1[16];
      char str2[16];
      strcpy(str1, "sssss");
      strcpy(str2, "kkkk");
      strcat(str1, str2);
      printf("%s", str1);
      return 0;
}

Output:

ssssskkkk

You have to allocate memory for your strings. In the above code, I declare str1 and str2 as character arrays containing 16 characters. I used strcpy to copy characters of string literals into them, and strcat to append the characters of str2 to the end of str1. Here is how these character arrays look like during the execution of the program:

After declaration (both are empty): 
str1: [][][][][][][][][][][][][][][][][][][][] 
str2: [][][][][][][][][][][][][][][][][][][][]

After calling strcpy (\0 is the string terminator zero byte): 
str1: [s][s][s][s][s][\0][][][][][][][][][][][][][][] 
str2: [k][k][k][k][\0][][][][][][][][][][][][][][][]

After calling strcat: 
str1: [s][s][s][s][s][k][k][k][k][\0][][][][][][][][][][] 
str2: [k][k][k][k][\0][][][][][][][][][][][][][][][]

Answer 3

strcat attempts to append the second parameter to the first. This won't work since you are assigning implicitly sized constant strings.

If all you want to do is print two strings out

printf("%s%s",str1,str2);

Would do.

You could do something like

char *str1 = calloc(sizeof("SSSS")+sizeof("KKKK")+1,sizeof *str1);
strcpy(str1,"SSSS");
strcat(str1,str2);

to create a concatenated string; however strongly consider using strncat/strncpy instead. And read the man pages carefully for the above. (oh and don't forget to free str1 at the end).

Answer 4

strcat concats str2 onto str1

You'll get runtime errors because str1 is not being properly allocated for concatenation

Answer 5

strcat(str1, str2) appends str2 after str1. It requires str1 to have enough space to hold str2. In you code, str1 and str2 are all string constants, so it should not work. You may try this way:

char str1[1024];
char *str2 = "kkkk";
strcpy(str1, "ssssss");
strcat(str1, str2);
printf("%s", str1);

Answer 6

The way it works is to:

1. Malloc memory large enough to hold copies of str1 and str2
2. Then it copies str1 into str3
3. Then it appends str2 onto the end of str3
4. When you're using str3 you'd normally free it free (str3);

Here's an example for you play with. It's very simple and has no hard-coded lengths. You can try it here: http://ideone.com/d3g1xs

See this post for information about size of char

#include <stdio.h>
#include <memory.h>

int main(int argc, char** argv) {

      char* str1;
      char* str2;
      str1 = "sssss";
      str2 = "kkkk";
      char * str3 = (char *) malloc(1 + strlen(str1)+ strlen(str2) );
      strcpy(str3, str1);
      strcat(str3, str2);
      printf("%s", str3);

      return 0;
 }