Joining 8 strings to form 1 string in C

Question

I am doing a C programming school project. In one part of the project, I need to join every 8 strings (each 4 characters in length) to form 1 string (each 32 characters in length).

For example, char *holdBinary[16] holds 16 strings such as:

holdBinary[0] = "0000";
holdBinary[1] = "0000";
holdBinary[2] = "0000";
holdBinary[3] = "0000";
holdBinary[4] = "0000";
holdBinary[5] = "0000";
holdBinary[6] = "0000";
holdBinary[7] = "0000";
holdBinary[8] = "0000";
holdBinary[9] = "0000";
holdBinary[10] = "0010";
holdBinary[11] = "1010";
holdBinary[12] = "0000";
holdBinary[13] = "0000";
holdBinary[14] = "0000";
holdBinary[15] = "0000";

I need to join every 8 strings in holdBinary to form 1 string which is held in holdRange char array. Thus, the results: holdeRange[0] = "00000000000000000000000000000000" which is formed by holdBinary[0], holdBinary[1] .... holdBinary[7] and holdeRange[1] = "00000000001010100000000000000000".

Here is the code:

char holdRange[2][33];  // I changed here and it works correct now


int hold = 0;
int z = 0;
int index2 = -1;

while(z<16)
{
    if(z%8 == 0)
    {

        index2++;
        strcpy(holdRange[index2] , holdBinary[z]);
    }
    else
    {

        strcat(holdRange[index2] , holdBinary[z]);
    }
    z++;

}
printf("%s" , holdRange[0]); --> prints 0000000000000000000000000000000000000000001010100000000000000000
printf("\n");
printf("%s" , holdRange[1]); --> prints 00000000001010100000000000000000

Thus, holdRange[0] is not equal to what must be. How can I fix it?

Answer 1

Using %s to printf something generally requires a \0 character at the end, which you did not leave room for in your string -- it needs to hold 33 characters if you want 32 "real" characters and a null terminator.

Answer 2

You allocate one block of memory for two strings. I assume that's because you confused yourself with the odd types you use. All you need is char* holdRange[2] and memory for each of the strings.

Sure, it could be made to work your way, but you're not thinking of \0 either.

Answer 3

If you just want to print out 32 characters and wants the characters stored in contiguous memory, give a format specifier to printf():

printf("%.32s" , holdRange[0]);
printf("\n");
printf("%.32s" , holdRange[1]);