浅析C语言编程中的数组越界问题

#include <stdio.h>

void f(int ai)
{
int aa[5]={1,2,3};
int i = 1;
for (i=0;i<10;i++)
 aa[i]=i;
printf("f()/n");
}

void main()
{
f(3);
printf("ok/n");
}

 

 

###sta.s###

     .file  "sta.c"                 ;说明汇编的源程序
    .section    .rodata           ;说明以下是只读数据区
.LC0:
    .string "f()"                  ;"f()" 的类型是string，地址为LC0
    .text                       ;代码段开始
.globl f                         ;f为全局可访问
    .type  f, @function            ; f是函数
f:
    pushl  %ebp
    movl  %esp, %ebp
    subl  $40, %esp
    movl  $0, -24(%ebp)
    movl  $0, -20(%ebp)
    movl  $0, -16(%ebp)
    movl  $0, -12(%ebp)
    movl  $0, -8(%ebp)
    movl  $1, -24(%ebp)
    movl  $2, -20(%ebp)
    movl  $3, -16(%ebp)
    movl  $1, -4(%ebp)
    movl  $0, -4(%ebp)
    jmp   .L2
.L3:
    movl  -4(%ebp), %edx
    movl  -4(%ebp), %eax
    movl  %eax, -24(%ebp,%edx,4)
    addl  $1, -4(%ebp)
.L2:
    cmpl  $9, -4(%ebp)
    jle   .L3
    movl  $.LC0, (%esp)
    call  puts
    leave
    ret
    .size  f, .-f                     ;用以计算函数f的大小
    .section    .rodata
.LC1:
    .string "ok"
    .text
.globl main
    .type  main, @function
main:
    leal  4(%esp), %ecx
    andl  $-16, %esp
    pushl  -4(%ecx)
    pushl  %ebp
    movl  %esp, %ebp
    pushl  %ecx
    subl  $4, %esp
    movl  $3, (%esp)
    call  f
    movl  $.LC1, (%esp)
    call  puts
    addl  $4, %esp
    popl  %ecx
    popl  %ebp
    leal  -4(%ecx), %esp
    ret
    .size  main, .-main
    .ident "GCC: (GNU) 4.1.2 20070115 (SUSE Linux)"        ;说明是用什么工具编译的
    .section    .note.GNU-stack,"",@progbits