a[0] is same as &a[0] and a (for 2 dim array) ?

[black_snail]

#include <stdio.h>
#define FORMAT "%d,%d\n"
main()
{
static int a[3][4]={1,3,5,7,9,11,13,15,17,19,21,23};
int *p;
A. for (p=a[0];p<a[0]+12;p++)
B. for (p=a[0];p<(&a[0])+12;p++)
C. for (p=a[0];p<a+12;p++)


{
if ((p-a[0])%4==0) printf("\n");
printf("%4d",*p);
}

}

A is working (Display the array value )
B and C not working , Why , I thought a[0] is same as &a[0] and a (for 2 dim array)

[dcyu]

a[0] is not same as &a[0] and a (for 2 dim array)
&a[0] is same as a ,but both of them are not same as a[0],
so y r wrong!

[black_snail]

printf("%o,%o,%o",&a[0],a[0],a) shows the same value , Why ?

[blh]

首先肯定的是a[0]、 &a[0]和 a的地址是相同的，只是随后的+12是不同的
a[0] + 12 是以a[0]地址为起始，按照a[0]的长度为单位+12
而&a[0]和 a只是以&a[0]和 a地址为起始，简单的将地址+12
这是编译器来决定的

[black_snail]

blh(老猫) , your answer is quite reasonable

[木棉花大榕树]

B. Because a[0] is a type int*, &a[0] is a type int **.
C. I think (*a)=a[0] not (*a)=a[0][0],so a+12=a[12],!=a[0][12]