[SOLVED] Segmentation fault in C code

[ranayehya]

Hello!
I want to concatenate 2 unsigned char (temp and counter) contain hexadecimal numbers and store them in another variable (after) but this code ives me segmentation fault. Could someone help me?
Thanks..

#include<stdio.h>
void concat()
{
    //unsigned char state[4][4];
    unsigned char temp[8] = {0x00  ,0x01  ,0x02  ,0x03  ,0x04  ,0x05  ,0x06,0x07};
    unsigned char i, j;
    unsigned char after[16] ={0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};
    unsigned char counter[8]  = {0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};
    // calculate the length of string s1
    // and store it in i
    for(i = 0; i <= 7; i++)
    {
        after[i] = temp[i];
    }
    for(i = 0; i <= 7; i++)
    {
        after[8+i] = temp[i];
    }
    for(i = 0; i <= 15; i++)
    {
        printf("%s\n", after[i]);
    }
   /* for(i=0;i<4;i++)
    {
        for(j=0;j<4;j++)
        {
            state[j][i] = counter[i*4 + j];
        }
    }*/
}
int main()
{
    concat();
    return 0;
}

 

[Easyrider83]

couter have size 8 bytes, but you exceed it with i = 3, j = 3 => i * 4 + j = 15
replace counter with 'after'

 

[ranayehya]

Hello, Easyrider83!
This piece of code is a comment. It is my bad that I did not delete it. My apologizes.

 

[Easyrider83]

printf("%s\n", after); -> this thing prints out null terminated string. Use %h or %H