gdb print array of soft array

Question

#include <stdio.h>
#include <malloc.h>
typedef struct _soft_array
{
    int len;
    int array[];//int array[0];
}SoftArray;

int main()
{  
    int i = 0;
    SoftArray* sa = (SoftArray*)malloc(sizeof(SoftArray) + sizeof(int) * 10);

    sa->len = 10;

    for(i=0; i<sa->len; i++)
    {
        sa->array[i] = i + 1;
    }

    for(i=0; i<sa->len; i++)
    {
        printf("%d\n", sa->array[i]);   
    }

    free(sa);

    return 0;
}

I use gdb to print the array of sa->array,then something confuse me:

(gdb) p sa->array
$1 = 0x602014
(gdb) p *sa->array@10
$2 = {[0] = 1, [1] = 2, [2] = 3, [3] = 4, [4] = 5, [5] = 6, [6] = 7, [7] = 8, [8] = 9, [9] = 10}
(gdb) p sizeof(int)
$3 = 4
(gdb) p *(char *)0x602014@40
$4 = "\001\000\000\000\002\000\000\000\003\000\000\000\004\000\000\000\005\000\000\000\006\000\000\000\a\000\000\000\b\000\000\000\t\000\000\000\n\000\000"
(gdb)

the first method went well (I add set print array-indexes on in .gdbinit).

sizeof(int)==4,then I use p *(char *)0x602014@40,due to my computer is little-endian，so 1 display with 01 00 00 00.

When display 7,the output become a?

How does that happen or I miss something ?

Are you perhaps debugging an optimized build? If so, the debugging experience is not reliable. — Jesper Juhl
– Jesper Juhl, Commented Nov 11, 2018 at 15:54

Some programmer dude · Accepted Answer · 2018-11-11 15:58:24Z

2

Note that the output isn't the character 'a', but '\a'. That is the alert character, in ASCII. Just like you later get '\b' and '\t' and '\n' for the ASCII equivalents of those characters.

This character escape reference contains a list of all standard C escape characters, together with their ASCII encoded values.

answered Nov 11, 2018 at 15:58

Some programmer dude

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1 Comment

52coder Over a year ago

not familiar with Escape sequences,reading the article learn much.command p *(char *)0x602014@40 just print every byte in octal,this is unusual for me.

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