019 一维数组 选择法对十个整数进行排序

/***********************019 一维数组***********************
* 使用选择法对十个整数进行排序。
* C语言精彩编程百例第19
*/

#include<stdio.h>
void main()
{
int i,j,min,temp;
int array[10];

printf("please input ten integer:\n");
for(i=0;i<10;i++)
{
printf("array[%d]=",i);
scanf("%d",&array[i]);
}
printf("The array is:");
for(i=0;i<9;i++)
{
min=i;
for(j=i;j<10;j++)
if(array[min]>array[j]) min =j;
temp=array[i];
array[i]=array[min];
array[min]=temp;
}
printf("\nThe result:\n");
for(i=0;i<10;i++)
printf("%d",array[i]);
printf("\n");
}

对应的汇编内容：

.file	"019.c"
.def	___main;	.scl	2;	.type	32;	.endef
.text
LC0:
.ascii "please input ten integer:\12\0"
LC1:
.ascii "array[%d]=\0"
LC2:
.ascii "%d \0"
LC3:
.ascii "The array is:\0"
LC4:
.ascii "\12The result:\12\0"
LC5:
.ascii "\12\0"
.align 2
.globl _main
.def	_main;	.scl	2;	.type	32;	.endef
_main:
pushl	%ebp
movl	%esp, %ebp
subl	$88, %esp
andl	$-16, %esp
movl	$0, %eax
movl	%eax, -76(%ebp)
movl	-76(%ebp), %eax
call	__alloca
call	___main
subl	$12, %esp         # printf("please input ten integer:\n")
pushl	$LC0
call	_printf
addl	$16, %esp
movl	$0, -12(%ebp)     # i=0
L4:
cmpl	$9, -12(%ebp)     # i<10
jle	L7
jmp	L5
L7:
subl	$8, %esp         # printf("array[%d]=",i);
pushl	-12(%ebp)
pushl	$LC1
call	_printf
addl	$16, %esp
subl	$8, %esp          # scanf("%d",&array[i]);
leal	-72(%ebp), %edx   # edx = &array[0]
movl	-12(%ebp), %eax   # eax = i
sall	$2, %eax	  # eax= eax*4
leal	(%eax,%edx), %eax # eax= &array[0]+i*4
pushl	%eax              # scanf("%d",&array[i]);
pushl	$LC2
call	_scanf
addl	$16, %esp
leal	-12(%ebp), %eax   # &i
incl	(%eax)            # i++
jmp	L4                # 下次循环
L5:
subl	$12, %esp         # printf("The array is:");
pushl	$LC3
call	_printf
addl	$16, %esp
movl	$0, -12(%ebp)     # i=0
L8:
cmpl	$8, -12(%ebp)     # i 和 9-1比较
jle	L11               # i<9
jmp	L9
L11:
movl	-12(%ebp), %eax
movl	%eax, -20(%ebp)   # min =i
movl	-12(%ebp), %eax
movl	%eax, -16(%ebp)   # j =i
L12:
cmpl	$9, -16(%ebp)     # j 和 10-1 比
jle	L15
jmp	L13
L15:
movl	-20(%ebp), %eax         # eax =min
movl	-16(%ebp), %edx         # edx =j
movl	-72(%ebp,%eax,4), %eax  # eax =[ebp+eax*4-72]=arrar[min]
cmpl	-72(%ebp,%edx,4), %eax  # array[min] 和 array[j] 比较
jle	L14                     # 小于等于跳
movl	-16(%ebp), %eax         # eax = j
movl	%eax, -20(%ebp)         # min =j
L14:
leal	-16(%ebp), %eax         # j++
incl	(%eax)
jmp	L12
L13:
movl	-12(%ebp), %eax           # eax =i
movl	-72(%ebp,%eax,4), %eax    # eax = array[i]
movl	%eax, -24(%ebp)           # temp=eax
movl	-12(%ebp), %edx           # edx=i
movl	-20(%ebp), %eax           # eax=min
movl	-72(%ebp,%eax,4), %eax    # eax=array[min]
movl	%eax, -72(%ebp,%edx,4)    # array[i]=eax
movl	-20(%ebp), %edx           # edx=min
movl	-24(%ebp), %eax           # eax=temp
movl	%eax, -72(%ebp,%edx,4)    # array[min]=eax
leal	-12(%ebp), %eax           # i++
incl	(%eax)
jmp	L8
L9:
subl	$12, %esp                 # printf("\nThe result:\n");
pushl	$LC4
call	_printf
addl	$16, %esp
movl	$0, -12(%ebp)             # i=0
L17:
cmpl	$9, -12(%ebp)             # i<10
jle	L20
jmp	L18
L20:
subl	$8, %esp                  # printf("%d",array[i]);
movl	-12(%ebp), %eax
pushl	-72(%ebp,%eax,4)
pushl	$LC2
call	_printf
addl	$16, %esp
leal	-12(%ebp), %eax          # i++
incl	(%eax)
jmp	L17                      # 下次循环
L18:
subl	$12, %esp
pushl	$LC5
call	_printf
addl	$16, %esp
leave
ret
.def	_scanf;	.scl	2;	.type	32;	.endef
.def	_printf;	.scl	2;	.type	32;	.endef

按照程序内容来看

 int i,j,min,temp;  的内存位置 -12(%epb)~-24(%epb)

int array[10];  的内存位置 [%ebp+0*4-72] ~ [%ebp+10*4-72]