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

    return (EXIT_SUCCESS);
 }

I want to concatenate the two strings, but it doesn't work.

You are failing to allocate a destination buffer of sufficient size, and attempting to write to non-modifiable memory. — David Heffernan, Aug 27 '13 at 14:29
When you say "it doesn't work?", what do you mean? What is it printing instead of what it's supposed to? Even if it's fairly obvious, you should also say what you think it's supposed to print, to make it clear. — qaphla, Aug 27 '13 at 14:29
see this question [Using strcat in C](http://stackoverflow.com/questions/419604/using-strcat-in-c) — lsalamon, Aug 27 '13 at 14:34
possible duplicate of [strcpy and strcat cause problems sometimes](http://stackoverflow.com/questions/7254648/strcpy-and-strcat-cause-problems-sometimes) — Binary Worrier, Aug 27 '13 at 14:53

score 42 · Accepted Answer · edited May 23 '17 at 11:55

42

The way it works is to:

Malloc memory large enough to hold copies of str1 and str2
Then it copies str1 into str3
Then it appends str2 onto the end of str3
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;
 }

edited May 23 '17 at 11:55

Community

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answered Aug 27 '13 at 14:36

dcaswell

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what's the difference of `str3 = "sssss"` and `strcpy(str3,str1)`? – BeCurious Aug 28 '13 at 00:55
str3 needs to point at memory that's large enough to hold the string. That's why I malloced it in the line above. – dcaswell Aug 28 '13 at 00:59
But after this code `char * str3 = (char *) malloc(1 +sizeof(char*) * (strlen(str1)+ strlen(str2)));` Why not use `str3 = "sssss"` but use `strcpy(str3,str1)` instead?:) – BeCurious Aug 28 '13 at 01:07
Because we don't want to point str3 at str1. It's a constant. There's no room to copy characters onto the end of it. – dcaswell Aug 28 '13 at 01:08
But you have malloced a memory,isn't it? – BeCurious Aug 28 '13 at 01:27
I malloced enough memory for copies of str2, str3, +1 more (for the string null termination character. – dcaswell Aug 28 '13 at 01:30
Multiplying by `sizeof(char)` is pointless (and not advised). – Nik Bougalis Aug 28 '13 at 01:43
@NikBougalis what do you mean?can you explain more detail? thanks :) – BeCurious Aug 28 '13 at 01:50
@BeCurious in C and C++ the result of `sizeof(char)` is always *guaranteed* to be 1. So multiplying by it is pointless. – Nik Bougalis Aug 28 '13 at 01:53
I also added a link to what sizeof char meant. – dcaswell Aug 28 '13 at 01:56
@dcaswell In this line `char * str3 = (char *) malloc(1 + strlen(str1)+ strlen(str2) );` . Do you need the `(char *) `? `malloc` confuses me when you get fancy. – cokedude Jan 20 '16 at 03:46
I get instant crash, why ? – Oct 31 '16 at 23:30
Isn't there a word missing in the step 4? – Jaroslav Záruba Feb 20 '17 at 09:11

kol · Answer 2 · 2013-08-27T14:41:13.103

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][][][][][][][][][][][][][][][]

score 3 · Answer 3 · answered Aug 27 '13 at 14:29

3

strcat concats str2 onto str1

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

answered Aug 27 '13 at 14:29

Sam I am says Reinstate Monica

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In fact `strcat` does return the concatenated string. – David Heffernan Aug 27 '13 at 14:29

score 2 · Answer 4 · answered Aug 27 '13 at 14:32

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.

score 1 · Answer 5 · answered Aug 27 '13 at 14:30

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);

score 1 · Answer 6 · answered Aug 27 '13 at 14:30

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).

What's the difference of `str1="ssss"` and `strcpy(str1,"ssss")`? — BeCurious, Aug 28 '13 at 00:56

Concatenate two char* strings in a C program

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