unexpected array modification

[j_hop_97]

I'm hoping someone can explain the behavior of the following program.
When I run it on my linux machine the value of "bit[0]" for the last
"for" loop is "0".
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define SIZE 4

int main(void)
{

int idx;
unsigned char bit[SIZE];

char tmpstr[1];

/* initialize the array*/
for (idx = 0;idx < SIZE; idx++) {
bit[idx] = (unsigned char)idx + 1;
}

printf("\n\n");
for (idx = 0;idx < SIZE; idx++) {
printf("%d. bit[%d] = '%d'\n", idx, idx, bit[idx]);
}

printf("\n\n");
for (idx = 0;idx < SIZE; idx++) {
printf("%d. bit[%d] = '%d'\n", idx, idx, bit[idx]);
}

strcpy(tmpstr, "0"); // why does this have any affect?

printf("\n\n");
for (idx = 0;idx < SIZE; idx++) {
printf("%d. bit[%d] = '%d'\n", idx, idx, bit[idx]);
}

return 0;
}

Makefile:

CC = gcc
CFLAGS = -Wall

all: test

test: test.c
$(CC) $(CFLAGS) -o $@ $<

clean:
rm -f *.o test

..PHONY: clean

[Bill Pursell]

j_hop_97 wrote:

I'm hoping someone can explain the behavior of the following program.
When I run it on my linux machine the value of "bit[0]" for the last
"for" loop is "0".
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define SIZE 4

int main(void)
{

int idx;
unsigned char bit[SIZE];

char tmpstr[1];

/* initialize the array*/
for (idx = 0;idx < SIZE; idx++) {

bit[idx] = (unsigned char)idx + 1;
}

printf("\n\n");
for (idx = 0;idx < SIZE; idx++) {
printf("%d. bit[%d] = '%d'\n", idx, idx, bit[idx]);
}

printf("\n\n");
for (idx = 0;idx < SIZE; idx++) {
printf("%d. bit[%d] = '%d'\n", idx, idx, bit[idx]);
}

strcpy(tmpstr, "0"); // why does this have any affect?

Because tmpstr only has a size of one, but you are writing two
characters to here, (a '0' and a '\0'). It sounds like
the '\0' is being written to bit[0]. Try making tmpstr have
size 2, and the problem will probably go away.

--
Bill Pursell

[Harald van Dijk]

j_hop_97 wrote:

I'm hoping someone can explain the behavior of the following program.
When I run it on my linux machine the value of "bit[0]" for the last
"for" loop is "0".

[...]

char tmpstr[1];

[...]

strcpy(tmpstr, "0"); // why does this have any affect?

This will try to store two characters in tmpstr: '0', and '\0'. That
final '\0' is the only way C standard library functions can know how
long a string is. You only have enough room for a single character, so
the '\0' gets stored in whatever happens to follow tmpstr in memory.
(You may even get the program to exit, or worse, especially if nothing
directly follows tmpstr in memory.)

[j_hop_97]

Harald van Dijk wrote:

j_hop_97 wrote:

I'm hoping someone can explain the behavior of the following program.
When I run it on my linux machine the value of "bit[0]" for the last
"for" loop is "0".

[...]

char tmpstr[1];

[...]

strcpy(tmpstr, "0"); // why does this have any affect?

This will try to store two characters in tmpstr: '0', and '\0'. That
final '\0' is the only way C standard library functions can know how
long a string is. You only have enough room for a single character, so
the '\0' gets stored in whatever happens to follow tmpstr in memory.
(You may even get the program to exit, or worse, especially if nothing
directly follows tmpstr in memory.)

Yep, that was it. Defining tmpstr as "char tmpstr[2]" solved the
problem. Thanks.

[Coos Haak]

Op 9 Sep 2006 09:10:07 -0700 schreef j_hop_97:

<snip>

/* initialize the array*/
for (idx = 0;idx < SIZE; idx++) {
bit[idx] = (unsigned char)idx + 1;

Lose the cast, it's unnecessary here, as most casts are.
--
Coos