Returning a string from a function in C with parameters

Question

I am trying to return a string from my sCopy() function, so I can print it in the main() function of my code. Please help.

const char *sCopy(char buffer[256], int i);

int main() {
    int i;
    int x;
    char buffer[256];
    char newBuffer;//[256];
    printf("Please enter a number: ");
    fgets(buffer, 256, stdin);
    i = atoi(buffer);

printf("The value you entered is %d.  Its double is %d.\n", i, i*2);
newBuffer = sCopy(buffer, i);  
printf(newBuffer);    
return 0;
}   

const char *sCopy(char buffer[256], int i){
    char nBuffer[256];
    char *t;        
    int x;
    for(x = 0; x < i; x++){
        strcat(t, buffer);
    }
    //t = nBuffer;
    return t;
}

whats the point to use for loop when u dont use index "x" at all? — someone_ smiley, Mar 15 '13 at 03:21
I using gcc, declarations are not allowed in the for loop, not unless you compile with flags. — kingsolo, Mar 15 '13 at 03:33

score 1 · Answer 1 · edited May 23 '17 at 10:32

My analysis below. From what I see here you should brush up on the difference between a pointer to char (char*) and a char array. I really appreciate though that you've tried to solve it yourself before asking.

const char *sCopy(char buffer[256], int i); 
/* let's start from here, what i represents? Keep in mind that the most of the   */
/* time an external developer will see just your declaration of a method, always */
/* try to give significant names to variables.                                   */

int main() {
    int i = 0;
    /* always initialize variables to default values, especially if they are     */
    /* going to be indexes in a buffer.                                          */  

    int x = 0;
    char buffer[256] = "";
    /* you can even initialize this to "" in order to mimic an empty string,     */
    /* that is a char array cointining just \0 (string null-terminator).         */

    char newBuffer[256] = "";
    /* same here, you always need to declare the size of a char array unless     */ 
    /* you initialize it like this -char newBuffer[] = "hello"-, in which case   */
    /* the size will automatically be 6 (I'll let you discover/understand */
    /* why 6 and not 5).                                                         */

    printf("Please enter a number: ");
    fgets(buffer, 256, stdin); // nice link at the bottom on input reading
    i = atoi(buffer);

    printf("The value you entered is %d.  Its double is %d.\n", i, i*2);
    newBuffer = sCopy(buffer, i);  
    printf(newBuffer);    
    return 0;
}   

/* I am not going to debate the idea of returning a char pointer here :)           */
/* Remember that in this case you are returning a pointer to some memory that has  */
/* been allocated somewhere inside your function and needs to be released (freed)  */
/* by someone outside your control. Are they going to remember it? Are they        */
/* going to do it? In this case "they" is "you" of course.                         */
/* I'll let you explore alternative approaches.                                    */

const char *sCopy(char buffer[256], int i){
    char nBuffer[256] = ""; // see above
    char *t = NULL;
    /* you always init a pointer to NULL. As you can see, initializing here will */
    /* make you think that there might be problem with the strcat below.           */

    int x; // ok here you can omit the init, but be aware of it.
    for(x = 0; x < i; x++){
        strcat(t, buffer);

    /* what are you missing here? this piece of code is supposed to concatenate the */
    /* input buffer to a brand new buffer, pointed by your variable t. In your implementation */
    /* t is just a pointer, which is nothing more than a number to a memory location */
    /* With the initialization, the memory location you are pointing to is NULL, which */
    /* if de-referenced, will cause massive problems.                                */
    /* What you are missing is the blank slate where to write your data, to which your */ 
    /* pointer will read from.                                                        */
    }
    //t = nBuffer;
   return t;
}

I really hope that this would help you. I am sorry but I can't write the solution just because I think it's better if you learn it the hard way. You can find plenty of tutorials of pointers to char and I am sure you will solve the problem.

(input reading) C scanf() and fgets() problem

score 0 · Answer 2 · answered Mar 15 '13 at 03:17

0

Memory for char *t is not allocated. You can implement the string copy directly using strdup.

char *newBuffer = NULL;
...
...
newBuffer = strdup(buffer);

answered Mar 15 '13 at 03:17

rashok

12,790
16
88
100

score 0 · Answer 3 · answered Mar 15 '13 at 03:18

Try

char* newBuffer; // point to a c-string // here the return value of sCopy

and

static char nBuffer[256]; // make nBuffer survive beyond function call

and just use n return nBuffer, instead of t.

This is one way you could do it, without allocations. Alternatively allocate memory and return it instead of static char.

score 0 · Answer 4 · edited May 23 '17 at 12:21

Try below:

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

void sCopy(char buffer[256], int i, char newBuffer[], int size)
{
    char *t;        
    for(int x = 0; x < i; x++){
      strcat(newBuffer, buffer);
     }
 }

int _tmain(int argc, _TCHAR* argv[])
{
   int i;
   char buffer[256] = {0};
   char newBuffer[256] = {0};
   printf("Please enter a number: ");
   fgets(buffer, 256, stdin);
   i = atoi(buffer);

   printf("The value you entered is %d.  Its double is %d.\n", i, i*2); 
   sCopy(buffer, i, newBuffer, 256);
   printf("%s", newBuffer);   
   return 0;
}

Note that fgets appends a new line character at the end of the input string, so if you input 6 then the string is 0x36 0x0a where 0x36 is the ASCII code for 6 and 0x0a is the ASCII code for new line character, to remove this new line character, see this link

score 0 · Answer 5 · answered Mar 15 '13 at 05:18

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

const char *sCopy(char buffer[256], int i);

int main() {
    int i;
    int x;
    char buffer[256];
    const char* newBuffer;//[256];
    printf("Please enter a number: ");
    fgets(buffer, 256, stdin);
    i = atoi(buffer);

printf("The value you entered is %d.  Its double is %d.\n", i, i*2);
newBuffer = sCopy(buffer, i);  //buffer tail '\n' need cut?
printf(newBuffer);

    free(newBuffer);
    return 0;
}   

const char *sCopy(char buffer[256], int i){
    char *t;        
    int x;

    t=(char*)malloc(strlen(buffer)*i + 1);
    *t='\0';
    for(x = 0; x < i; x++){
        strcat(t, buffer);
    }
    //t = nBuffer;
    return (const char*)t;
}

Returning a string from a function in C with parameters

5 Answers5