char array not displaying when passed to a function

Question

Why is the passing of char array not showing? The location of the pointer is passed to a function.

char plaintext[] = {
         0xCD,  0x76,  0x43,  0xF0,
         0x72,  0xA4,  0xA0,  0x82,
}

Given

 void displayArray(char** plaintext, int size) {
 // int newSize = sizeof(**plaintext);
 int i;

 for(i=0; i < size; ++i) {
     printf("%02X ", (0xff & *plaintext[i]));
    if (((i+1)% 8) == 0)    // as index starts from 0, (i+1)
        printf("\n");
 }
}

in main()

        char* plaintextCopy;
        plaintextCopy = (char*) malloc(numberOfItems*sizeof(char));

        memcpy(plaintextCopy, plaintext, numberOfItems);

        displayArray(&plaintextCopy, numberOfItems);

How are you setting numberOfItems? I don't think memcpy() and your displayArray() use the same definition. — John C
– John C, Commented Mar 23, 2014 at 19:54
int numberOfItems = sizeof(plaintext); printf("Number of items: %d \n", numberOfItems); — Moirisa Dikaiosýni
– Moirisa Dikaiosýni, Commented Mar 23, 2014 at 20:01

Sebastian-Laurenţiu Plesciuc · Accepted Answer · 2014-03-23 20:08:34Z

1

Based on your code :

void displayArray(char** plaintext, int size)
{
    int i;

    for(i=0; i < size; i++) 
    {
        printf("%02X ", (0xff & (*plaintext)[i]));
        if(((i+1)% 8) == 0)    // as index starts from 0, (i+1)
            printf("\n");
    }
}

int main(void)
{


    char plaintext[] = {
         0xCD,  0x76,  0x43,  0xF0,
         0x72,  0xA4,  0xA0,  0x82,
    };

    int numberOfItems = sizeof(plaintext);

     char* plaintextCopy;
     plaintextCopy = (char*) malloc(numberOfItems*sizeof(char));

     memcpy(plaintextCopy, plaintext, numberOfItems);

     displayArray(&plaintextCopy, numberOfItems);

    return 0;
}

It outputs :

CD 76 43 F0 72 A4 A0 82

Also, if you're sending an array that you want to display or change the values of, you don't need to send a double pointer to a function. A regular pointer would do. You should only use double pointers if the original array changes it's location in memory, that is, it's getting a new pointer after the function returns.

edited Mar 23, 2014 at 20:08

answered Mar 23, 2014 at 19:50

Sebastian-Laurenţiu Plesciuc

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4 Comments

Sebastian-Laurenţiu Plesciuc Over a year ago

try with this modification.

Sebastian-Laurenţiu Plesciuc Over a year ago

Ok. I have tested the code and I will include it fully in the answer shortly.

Moirisa Dikaiosýni Over a year ago

What if I want to calculate the size in the call function? eg. ` void displayArray(char** plaintext) { int i; int size = sizeof(*plaintext);` ?

Sebastian-Laurenţiu Plesciuc Over a year ago

It will return 4 since that's the size of (char *) i.e. a pointer to char. sizeof() only works on char[]. You can either keep the length in a separate variable or hold it in the array at the start or at the end. Or you can use a 0x00 to show that it's the end of the array.

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