CODEWITHSUNDEEP
cstringconcatenation
user476145
user476145

Reputation: 445

C string concatenation

I am trying to input 2 strings in C, and output a 3rd string which the concatenation of string 1 and 2.

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

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

    char stringarray1 [30];
    char stringarray2 [30];
    char stringarray3 [30];

    int length;

    printf("Please enter some text less than 30 characters long\n");
    scanf("%[a-z, ]", &stringarray1);

    printf("Please enter some text less than 30 characters long\n");
    scanf("%[a-z, ]", &stringarray2);

    strcat(stringarray1, stringarray2);


    //length = strlen(stringarray);
    printf("The combined string is %s\n", stringarray1);


    return (EXIT_SUCCESS);
}

It allows me to input the first string, but then prints the seond and third printf statement, without allowing me to enter the second string.

How can i enter the second string without it exiting? Why does it exit?

Upvotes: 1

Views: 3214

Answers (5)

user411313
user411313

Reputation: 3990

You know the difference between an chararray a stringarray? I think not. Also you must allocate enough space for you result-string, like:

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

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

    char chararray1 [31];
    char chararray2 [16];
    char chararray3 [30];

    int length;

    printf("Please enter some text less than 15 characters long\n");
    scanf("%15[a-z, ]", stringarray1);

    printf("Please enter some text less than 15 characters long\n");
    scanf("%15[a-z, ]", stringarray2);

    strcat(stringarray1, stringarray2);


    //length = strlen(stringarray);
    printf("The combined string is %s\n", stringarray1);


    return (EXIT_SUCCESS);
}

Upvotes: 1

T.E.D.
T.E.D.

Reputation: 44794

At a guess, I'd say you need a \n inside your scanf specification strings. You should probably also be asking for %s instead of %[a-z, ]. The way you have it a single character would match, then when the second scanf is hit, the next character would match.

Also there are some safety issues. You should probably be checking to make sure both strings together aren't larger than 30 chars. As Benoit pointed out, arrays in C also double as pointers to the first character, so you don't use the address-of operator (&) on them when passing them to the C IO routines. That would give the routine the address of your array pointer itself, which the IO routine would then merrily wipe out with character data. Not good!

Upvotes: 1

Benoit Thiery
Benoit Thiery

Reputation: 6387

You should not put the & for your arrays that are already pointers.

scanf("%[a-z, ]", stringarray2);

Upvotes: 2

AndersK
AndersK

Reputation: 36082

I would recommend using fgets instead of scanf, that you can specify a max length (buffer target size - 1) and avoid surprises.

also the target buffer should be twice the size of your input buffers otherwise you risk a memory overwrite

Upvotes: 0

Flinsch
Flinsch

Reputation: 4341

Have a look at the documentation of scanf: Your format string "%[a-z, ]" does not what you seem to think what it does. Just use "%s" for strings instead. (It's without any character checks).

Upvotes: 6

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