CODEWITHSUNDEEP
c
John Nyingi
John Nyingi

Reputation: 1110

Return a Char array in C

I am building a cipher program but i can't figure out how to return an array of char[] my cipher method

char *cipherinput(int ciphercount){
int i=0;
char *cipher[MAX_CIPHER_SIZE];


if(ciphercount>=2){
    printf("Enter the Cipher!\n");
    //loop through and add
    for(i=0;i<ciphercount;i++){
        scanf(" %c", cipher[i]);
    }
}

   return cipher;
}

my main method has

#define MAX_CIPHER_SIZE 16
#define MAX_TEXT_SIZE 256
int main()
{
   int ciphercount=0, textcount=0,i=0,j=0,k=0;
   int *cipher_, *text_, N=0,N_;

   printf("Enter size of Cipher!\n");
   scanf("%d", &ciphercount);

   if(ciphercount>=2){
    cipher_ = cipherinput(ciphercount);
   }
   else{
     printf("Start again / Cipher size should be greater or equal to 2\n");
     main();
   }
   return 0;
 }

I've tried several methods such as char** (string) with no success.

Upvotes: 2

Views: 5185

Answers (3)

sg7
sg7

Reputation: 6298

If you want to use char *cipher outside your function 

char *cipherinput(int ciphercount){
     int i=0;
     char *cipher[MAX_CIPHER_SIZE];
     //...
     return *cipher;
}

You need to allocate memory for your cipher buffer. Your cipher[MAX_CIPHER_SIZE] will not exist once the cipherinput function returns. This is the most common approach:

 char* cipherinput(int ciphercount) {
        // ...
        char *cipher = malloc( sizeof(char) * (MAX_CIPHER_SIZE+1) ); // +1 to guarantee null termination.

        //...
        return cipher;
    }

This approach requires you to remember to free the memory allocated for cipher once you do not needed.

In embedded systems you may want to avoid memory allocation, since allocation and deallocation may not be timely operation. In such case, the best approach is to pass a buffer to your function: 

char cipher[MAX_CIPHER_SIZE+1];    // buffer outside the function
cipherinput(ciphercount, cipher);  // function call

Function implementation:

void cipherinput(int ciphercount, char * cipher){
 // access elements of cipher buffer via cipher[i]);
 // ...
}

Upvotes: 1

selbie
selbie

Reputation: 104474

You are returning a pointer to stack memory, which is undefined behavior. Most likely your returned string will be corrupted shortly after the function returns or after the invocation of another function.

This is closer to what you want:

char* cipherinput(int ciphercount) {
    int i=0;
    char cipher[MAX_CIPHER_SIZE+1]; // +1 to guarantee null termination.
    cipher[0] = '\0';

    if(ciphercount>=2){
        printf("Enter the Cipher!\n");
        //loop through and add
        for(i=0;i<ciphercount;i++){
            scanf(" %c", cipher[i]);
        }
        cipher[ciphercount] = '\0'; // null terminate
    }

    return strdup(cipher);   // this is the same as ptr=malloc(strlen(cipher+1)) followed by strcpy(ptr,cipher)
}

The function returns a copy of the string the user typed in. The caller of this function is expected to invoke free on the returned pointer after it's done with it. If ciphercount < 2, the function will return an empty string.

Upvotes: 4

pm100
pm100

Reputation: 50110

Normal way 1, Allocate result in function and return it

    char *cipherinput(int ciphercount){
      char *cipher = malloc(MAX_CIPHER_SIZE);
...
      return cipher;
    }

Note that your caller will have to free the result

Normal way 2, Caller creates and passes result buffer

void cipherinput(int ciphercount, char * cipher){
 ....
}

caller goes

    char cipher[MAX_CIPHER_SIZE];
    cipherinput(x, cipher);

Normal way #3. HAve fixed static buffer in function. NOTE THIS IS NOT rentrant or thread safe. 

char *cipherinput(int ciphercount){
  static char cipher[MAX_CIPHER_SIZE];
  .....
  return cipher;
}

In general #2 is probably best since you are less likely to leak

Upvotes: 1

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