how can i concatenate an int with a char * in C?

Question

i know how to convert from an int to a "string." but then, i don't know how to concatenate that "string" to a char *. this is the code i have so far

char * result = "abcdefg";
char dummy[10];
int x = 12;
sprintf(dummy, "%d", x);

i have tried to print "dummy" and it worked. but when i used the function "strcat(result, dummy)" and tried to print "result", it gave me segmentation fault. please help me

Answer 1

The previous responders have the right idea but the wrong target!



Instead, try

char result[32] = "abcdefg";

char dummy[10];

int x=12;

sprintf(dummy, "%d", x);

strcat(result, dummy);



What you were doing was not allocating space for result! Your compiler just gave it enough space for its initializing string, and not enough for concatenating anything onto it.

Answer 2

You need to make sure your output string is big enough to hold the concatenated string (length of result string + length of int converted to string). dummy only has space for 9 characters + the "null character" (\0) that's why you get a segfault.



If you are using sprintf() to convert your int, you can do the concatenation at the same time:



char dummy[1024]; /* or something big enough */

sprintf(dummy, "%s%d", result, x);

printf("%s\n", dummy); /* should output abcdefg12 */

Answer 3

define a larger array for dummy and try again. It seems that it is very small.



abcdefg is 7 chars and plus the NULL char ( '\0' ) that makes 8, you have 2 left!!



make it



char dummy[100] or something.



also read the below link about strcat..



Peace

Answer 4

First you have to put an index into the array to view its content for example:



for(int indx = 0; indx < 10; indx++)

{

sprintf("%c",dummy[indx],"%d",x);

}



indx is incremented the caracter "%c" display the caracter content of the array and you dont need to convert to int.