Append string to char array

Question

I have a code with a character array as follows:

int main (int argc, char** argv)
{
   char arr[] = "%55u%10$n";
   return 0;
}

I wish to dynamically set 10 through a string input from the user, for example argv[1].

What would the correct syntax of the following be:

char arr[] = "%55u%" argv[2] "$n"; // which would basically be "%55u%10$n" if argv[1] == "10"

That's not the number 10. That's the two characters '1' and '0'. — Ken White
– Ken White, Commented Dec 3, 2019 at 1:44
You cannot rely on concatenation of string literals with a variable not known until runtime. The compiler has no idea what argv[1] is going to be. See Remy's answer or use strcat — David C. Rankin
– David C. Rankin, Commented Dec 3, 2019 at 1:50

Remy Lebeau · Accepted Answer · 2019-12-03 02:18:48Z

2

You can use sprintf():

#include <stdio.h>

int main (int argc, char** argv)
{
   if (argc > 1)
   {
      char arr[20];
      sprintf(arr, "%%55u%%%.12s$n", argv[1]);
      ...
   }
   return 0;
}

Alternatively, use strcpy() and strcat() (use with care!):

#include <string.h>

int main (int argc, char** argv)
{
   if (argc > 1)
   {
      char arr[20];
      strcpy(arr, "%55u%");
      strcat(arr, argv[1]);
      // or: strcat_s(arr, sizeof(arr)-3, argv[1]);
      strcat(arr, "$n");
      ...
   }
   return 0;
}

edited Dec 3, 2019 at 2:18

answered Dec 3, 2019 at 1:48

Remy Lebeau

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Comments

ggorlen · Accepted Answer · 2019-12-03 01:50:12Z

1

Use sprintf. Be sure to escape literal % characters and ensure your buffer is large enough to hold the result string.

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

int main (int argc, char **argv) {
    if (argc < 2) { return 0; }

    char arr[strlen(argv[1])+8];
    sprintf(arr, "%%55u%%%s$n", argv[1]);
    printf("%s\n", arr);

    return 0;
}

answered Dec 3, 2019 at 1:50

ggorlen

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6 Comments

Aykhan Hagverdili Over a year ago

Using a VLA may not be the best option though

ggorlen Over a year ago

Maybe not, but it depends on the use case. Hardcoding 20 might not be the best option, either. Can you elaborate? If a huge input is provided, the stack frame will blow up--would this be a security risk?

Remy Lebeau Over a year ago

I hard-coded 20 in my example because I originally used %d for sprintf(), and the max value for a 32-bit integer is 10 digits. Then I saw the OP's desire to use argv for user input,so I switched my example to %s instead, but 20 is a good limit unless the OP needs more digits

ggorlen Over a year ago

Although OP chose "10", it's not entirely clear that they want to limit input to digits otherwise, and there's nothing preventing the user from providing some non-integer string as it stands.

Aykhan Hagverdili Over a year ago

@ggorlen what about dynamically allocating memory?

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