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一个很有深度的C++内存问题---GDB调试(一)

發(fā)布時間:2023/12/9 c/c++ 33 豆豆
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先看代碼(a.c):

#include <stdio.h>int main(int argc, char* argv[]) {int a[4]={1,2,3,4};int *ptr1=(int*)(&a+1);int *ptr2=(int*)((int)a+1);int *ptr3=(int*)(a+1);printf("a = %x, &a = %x, &a+1 = %x, (int)a+1 = %x\n", a, &a, &a+1, (int)a+1);printf("ptr1 = %x, ptr2 = %x, ptr3 = %x, &ptr1[-1] = %x\n", ptr1, ptr2, ptr3, &ptr1[-1]);printf("ptr1[-1] = %x, *ptr2 = %x, *ptr3 = %x\n",ptr1[-1],*ptr2, *ptr3);return 0; }


再看調(diào)試情況:

[zcm@t #81]$make clean rm -f *.o a [zcm@t #82]$make gcc -g -c -o a.o a.c gcc -g -o a a.o [zcm@t #83]$gdb a GNU gdb (GDB) Fedora (7.0-3.fc12) Copyright (C) 2009 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html> This is free software: you are free to change and redistribute it. There is NO WARRANTY, to the extent permitted by law. Type "show copying" and "show warranty" for details. This GDB was configured as "i686-redhat-linux-gnu". For bug reporting instructions, please see: <http://www.gnu.org/software/gdb/bugs/>... Reading symbols from /mnt/hgfs/Linux_Share/t/a...done. (gdb) list 1 #include <stdio.h> 2 3 int main(int argc, char* argv[]) 4 { 5 int a[4]={1,2,3,4}; 6 int *ptr1=(int*)(&a+1); 7 int *ptr2=(int*)((int)a+1); 8 int *ptr3=(int*)(a+1); 9 printf("a = %x, &a = %x, &a+1 = %x, (int)a+1 = %x\n", a, &a, &a+1, (int)a+1); 10 printf("ptr1 = %x, ptr2 = %x, ptr3 = %x, &ptr1[-1] = %x\n", ptr1, ptr2, ptr3, &ptr1[-1]); (gdb) list 11 printf("ptr1[-1] = %x, *ptr2 = %x, *ptr3 = %x\n",ptr1[-1],*ptr2, *ptr3); 12 13 return 0; 14 }(gdb) b 5 Breakpoint 1 at 0x80483ce: file a.c, line 5. (gdb) b 11 Breakpoint 2 at 0x8048472: file a.c, line 11. (gdb) b 13 Breakpoint 3 at 0x80484a0: file a.c, line 13. (gdb) run Starting program: /mnt/hgfs/Linux_Share/t/a Breakpoint 1, main (argc=1, argv=0xbffff2f4) at a.c:5 5 int a[4]={1,2,3,4}; Missing separate debuginfos, use: debuginfo-install glibc-2.11-2.i686 (gdb) c Continuing. a = bffff214, &a = bffff214, &a+1 = bffff224, (int)a+1 = bffff215 ptr1 = bffff224, ptr2 = bffff215, ptr3 = bffff218, &ptr1[-1] = bffff220Breakpoint 2, main (argc=1, argv=0xbffff2f4) at a.c:11 11 printf("ptr1[-1] = %x, *ptr2 = %x, *ptr3 = %x\n",ptr1[-1],*ptr2, *ptr3); (gdb) c Continuing. ptr1[-1] = 4, *ptr2 = 2000000, *ptr3 = 2Breakpoint 3, main (argc=1, argv=0xbffff2f4) at a.c:13 13 return 0; (gdb) x /24xb a 0xbffff214: 0x01 0x00 0x00 0x00 0x02 0x00 0x00 0x00 0xbffff21c: 0x03 0x00 0x00 0x00 0x04 0x00 0x00 0x00 0xbffff224: 0x24 0xf2 0xff 0xbf 0x15 0xf2 0xff 0xbf (gdb)


最后看圖解:

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