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[Bug 1894029] [NEW] qemu-i386 malloc error
From: |
Tony.LI |
Subject: |
[Bug 1894029] [NEW] qemu-i386 malloc error |
Date: |
Thu, 03 Sep 2020 04:12:10 -0000 |
Public bug reported:
Hi!I use qemu-i386-static on 64 bit machines.And memory request succeeded, but
the pointer is wrong.
This is my test program:
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, char **argv)
{
void *pa=0,*pb=0,*pc=0,*pd=0;
pa = malloc(sizeof(uint32_t));
pb = malloc(sizeof(uint32_t));
pc = malloc(4);
pd = malloc(4);
printf("pa: 0x%x\n",pa);
printf("pb: 0x%x\n",pb);
printf("pc: 0x%x\n",pc);
printf("pd: 0x%x\n",pd);
printf("uint32_t:%d\n",sizeof(uint32_t));
free(pa);
free(pb);
free(pc);
free(pd);
return 0;
}
And it is wrong:
pa: 0x400051a0
pb: 0x400051b0
pc: 0x400051c0
pd: 0x400051d0
uint32_t:4
Why did I apply for 4 bytes of space, but the pointer only increased by 2
bytes??
Is it a BUG??
** Affects: qemu
Importance: Undecided
Status: New
--
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https://bugs.launchpad.net/bugs/1894029
Title:
qemu-i386 malloc error
Status in QEMU:
New
Bug description:
Hi!I use qemu-i386-static on 64 bit machines.And memory request succeeded,
but the pointer is wrong.
This is my test program:
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, char **argv)
{
void *pa=0,*pb=0,*pc=0,*pd=0;
pa = malloc(sizeof(uint32_t));
pb = malloc(sizeof(uint32_t));
pc = malloc(4);
pd = malloc(4);
printf("pa: 0x%x\n",pa);
printf("pb: 0x%x\n",pb);
printf("pc: 0x%x\n",pc);
printf("pd: 0x%x\n",pd);
printf("uint32_t:%d\n",sizeof(uint32_t));
free(pa);
free(pb);
free(pc);
free(pd);
return 0;
}
And it is wrong:
pa: 0x400051a0
pb: 0x400051b0
pc: 0x400051c0
pd: 0x400051d0
uint32_t:4
Why did I apply for 4 bytes of space, but the pointer only increased by 2
bytes??
Is it a BUG??
To manage notifications about this bug go to:
https://bugs.launchpad.net/qemu/+bug/1894029/+subscriptions
- [Bug 1894029] [NEW] qemu-i386 malloc error,
Tony.LI <=