Printing error while accessing consecutive memory locations

Question

Following is a code to see how different data types are stored in memory.

#include <stdio.h>

void newline(void)
{
    putchar('\n');
}

void showbyte(char *string, int len)
{
    int i;
    for (i = 0; i < len; i++)
    printf("%p\t0x%.2x\n", string+i, *(string+i));
}

int main()
{
    int i = 12345;
    float f = 1234.5;
    double d = 1234.5;
    char name[] = "12345";
    showbyte((char *)&i, sizeof i);
    newline();
    showbyte((char *)&f, sizeof f);
    newline();
    showbyte((char *)&d, sizeof d);
    newline();
    showbyte((char *)&name, sizeof name);
    return 0;
}

Output

0x7fff8a9ab2cc  0x39
0x7fff8a9ab2cd  0x30
0x7fff8a9ab2ce  0x00
0x7fff8a9ab2cf  0x00

0x7fff8a9ab2c8  0x00
0x7fff8a9ab2c9  0x50
0x7fff8a9ab2ca  0xffffff9a
0x7fff8a9ab2cb  0x44

0x7fff8a9ab2c0  0x00
0x7fff8a9ab2c1  0x00
0x7fff8a9ab2c2  0x00
0x7fff8a9ab2c3  0x00
0x7fff8a9ab2c4  0x00
0x7fff8a9ab2c5  0x4a
0x7fff8a9ab2c6  0xffffff93
0x7fff8a9ab2c7  0x40

0x7fff8a9ab2b0  0x31
0x7fff8a9ab2b1  0x32
0x7fff8a9ab2b2  0x33
0x7fff8a9ab2b3  0x34
0x7fff8a9ab2b4  0x35
0x7fff8a9ab2b5  0x00

The IEEE-754 representation for float 1234.5 is 0x449a5000, and for double 1234.5 is 0x40934A0000000000. When it printed the float and double variable contents, it shows a 4-byte content.

ie, 0x7fff8a9ab2ca 0xffffff9a and 0x7fff8a9ab2c6 0xffffff93.

But each memory location can store only 1-byte of data, then why does it happen?

@EricPostpischil Thank you.. I didn't find that link before... — noufal, Oct 09 '13 at 12:07

score 2 · Accepted Answer · edited May 23 '17 at 10:25

unsigned is your friend in this case, because char is by default a signed type

if you change some of the types from char to unsigned char you get the correct result:

#include <stdio.h>

void newline(void)
{
    putchar('\n');
}

void showbyte(unsigned char *string, int len)
{
    int i;
    for (i = 0; i < len; i++)
      printf("%p\t0x%.2x\n", string+i, *(string+i));
}

int main()
{
    int i = 12345;
    float f = 1234.5;
    double d = 1234.5;
    char name[] = "12345";
    showbyte((unsigned char *)&i, sizeof i);
    newline();
    showbyte((unsigned char *)&f, sizeof f);
    newline();
    showbyte((unsigned char *)&d, sizeof d);
    newline();
    showbyte((unsigned char *)&name, sizeof name);
    return 0;
}

In this case you get:

0x7fff5d5a98b8  0x39
0x7fff5d5a98b9  0x30
0x7fff5d5a98ba  0x00
0x7fff5d5a98bb  0x00

0x7fff5d5a98b4  0x00
0x7fff5d5a98b5  0x50
0x7fff5d5a98b6  0x9a
0x7fff5d5a98b7  0x44

0x7fff5d5a98a8  0x00
0x7fff5d5a98a9  0x00
0x7fff5d5a98aa  0x00
0x7fff5d5a98ab  0x00
0x7fff5d5a98ac  0x00
0x7fff5d5a98ad  0x4a
0x7fff5d5a98ae  0x93
0x7fff5d5a98af  0x40

0x7fff5d5a98a2  0x31
0x7fff5d5a98a3  0x32
0x7fff5d5a98a4  0x33
0x7fff5d5a98a5  0x34
0x7fff5d5a98a6  0x35
0x7fff5d5a98a7  0x00

0x9a and 0x93 are negative numbers :-) Maybe the compiler upgraded this value to integer or printf did it.. — pfmaggi, Oct 09 '13 at 05:38
yes, printf is doing that job, i could see it from this code..`int main() {char i = -3; printf("%.2x\n", i); return 0;}`. But why does it avoid `%.2x`? — noufal, Oct 09 '13 at 06:11

Printing error while accessing consecutive memory locations

1 Answers1