栈再次理解

不好意思，还有一个补充：
代码如下：
#include <stdio.h>

void a()
{
    int i;
    int j;

    printf( "i's address is %p\n", &i );
    printf( "j's address is %p\n", &j );
}

void b()
{
    int x;
    int y;

    printf( "x's address is %p\n", &x );
    printf( "y's address is %p\n", &y );

}

int main( void )
{
    a();
    b();
    return( 0 );
}


输出为：
i's address is 0xbfc8bd24
j's address is 0xbfc8bd20
x's address is 0xbfc8bd24
y's address is 0xbfc8bd20
提高内存利用率，一个调用完，调另一个。

大家好：
      有这样一个问题和我想象的不一样，问题如下：
代码：
#include <stdio.h>

int main( void )
{
    int i;
    int j;
    int a[10];

    printf( "i's address is %p\n", &i );
    printf( "j's address is %p\n", &j );
    printf( "a's address is %p\n", a );

    for( i = 0; i < 10; i++ )
    {
        printf( "a[%d]'s address is %p\n", i, &a[i] );
    }

    return( 0 );
}


输出如下：
i's address is 0xbfa76f60
j's address is 0xbfa76f5c
a's address is 0xbfa76f34
a[0]'s address is 0xbfa76f34
a[1]'s address is 0xbfa76f38
a[2]'s address is 0xbfa76f3c
a[3]'s address is 0xbfa76f40
a[4]'s address is 0xbfa76f44
a[5]'s address is 0xbfa76f48
a[6]'s address is 0xbfa76f4c
a[7]'s address is 0xbfa76f50
a[8]'s address is 0xbfa76f54
a[9]'s address is 0xbfa76f58

我想的是a[0]的地址应该是0xbfa76f58，a[1] 的是0xbfa76f54，。。。。。，但是事实上不是这样刚好相反。是不是进入栈中的顺序是a[10]先进，然后依次进入，还是和编译器有关系，请大家互相讨论，各抒己见，我学习一下，哈哈，谢谢。