memcpy erase variables in char array in C

Question

I create multi dimensioanl array and write it to console

char a[5][10];

strcpy(a[0], "111111");
strcpy(a[1], "211112");
strcpy(a[2], "311113");
strcpy(a[3], "411114");
strcpy(a[4], "511115");


printf("size : %d \n", sizeof(a));

int i;
for(i = 0; i < 5; i++)
{
    printf("%d : %s \n",i,a[i]);
}

result is

size : 50 
0 : 111111 
1 : 211112 
2 : 311113 
3 : 411114 
4 : 511115 

then I copy the array to another array, and write both of them to console

char a[5][10];
char b[][10]={"0"};

strcpy(a[0], "111111");
strcpy(a[1], "211112");
strcpy(a[2], "311113");
strcpy(a[3], "411114");
strcpy(a[4], "511115");

memcpy(&b,&a,sizeof(a));

printf("sizeof(a) : %d \n", sizeof(a));
int i;
for(i = 0; i < 5; i++)
{
    printf("%d : %s \n",i,a[i]);
}

printf("sizeof(b) : %d \n", sizeof(b));
for(i = 0; i < 5; i++)
{
    printf("%d : %s \n",i,b[i]);
}

result is:

sizeof(a) : 50 
0 :  
1 :  
2 :  
3 :  
4 : 511115 
sizeof(b) : 10 
0 : 111111 
1 : 211112 
2 : 311113 
3 : 411114 
4 : 511115 

what happened variables in a array ? why contents of a array are empty ? I use Ubuntu 14.04 and gcc version is (Ubuntu 4.8.4-2ubuntu1~14.04.1) 4.8.4

Answer 1

The size of b is too small to store all contents of a, so you accessed out of the allocated memory, which is very dangerous.

Do allocate enough memory.

char a[5][10];
char b[5][10]={{"0"}}; /* change this line */

strcpy(a[0], "111111");
strcpy(a[1], "211112");
strcpy(a[2], "311113");
strcpy(a[3], "411114");
strcpy(a[4], "511115");

memcpy(&b,&a,sizeof(a));

printf("sizeof(a) : %d \n", (int)sizeof(a)); /* change this line to pass data having correct type */
int i;
for(i = 0; i < 5; i++)
{
    printf("%d : %s \n",i,a[i]);
}

printf("sizeof(b) : %d \n", (int)sizeof(b)); /* change this line to pass data having correct value */
for(i = 0; i < 5; i++)
{
    printf("%d : %s \n",i,b[i]);
}

This code may show why the contents in a seems deleted: 0x00 is written to a[i][0] and printf() interpreted them as end of string.

#include <stdio.h>

int main(void) {
    char a[5][10];
    char b[][10]={"0"};

    strcpy(a[0], "111111");
    strcpy(a[1], "211112");
    strcpy(a[2], "311113");
    strcpy(a[3], "411114");
    strcpy(a[4], "511115");

    memcpy(&b,&a,sizeof(a));

    int i, j;
    for (i = 0; i < 5; i++)
    {
        for (j = 0; j < 10; j++) printf("%02X ", (unsigned int)(unsigned char)a[i][j]);
        putchar('\n');
    }

    return 0;
}

Answer 2

The memcpy overwrote the contents of the array a. Why did it do that?

The array b was allocated only 10 bytes, because char b[][10] means "allocate space for an array of 10-char strings, the number of strings being specified by the initializer." In your case, the initializer is "0", ie, one string. So it allocated space for one 10-char string.

Since the whole thing is on the stack, the layout is reversed. Thus, the location of b comes first, and then a.

So, when you write into b, it starts correctly, writes the first string. When you write the second string, it overflows into the array a's space and kills it. 4 times. The last string survives.