[英]Pass a pointer as array in function in C?
void f(int *a, int n)
{
int i;
for (i = 0; i < n; i++)
{
printf("%d\n", *(a+i));
}
}
The above code worked ok if in main()
I called: 如果在
main()
我调用了上面的代码就可以了:
int a[] = {1,2,3};
int *p = a;
f(a, 3);
But if in main()
, I did: 但如果在
main()
,我做了:
int *a =(int*) {1,2,3};
f(a, 3);
Then, the program will crash. 然后,程序将崩溃。 I know this might look weird but I am studying and want to know the differences.
我知道这可能看起来很奇怪,但我正在研究并希望了解其中的差异。
It's because of the cast. 这是因为演员。 This line says:
这条线说:
int *a =(int*) {1,2,3};
Treat the array {1,2,3}
as a pointer to an int. 将数组
{1,2,3}
视为指向int的指针。 On a 32 bit machine, the value of the pointer is now 1
, which is not what you want. 在32位机器上,指针的值现在为
1
,这不是您想要的。
However, when you do: 但是,当你这样做时:
int *p = a;
The compiler knows that it can decay the array name to a pointer to it's first element. 编译器知道它可以将数组名称衰减为指向它的第一个元素的指针。 It's like you'd actually written:
这就像你实际上写的:
int *p = &(a[0]);
Similarly, you can just pass a
straight in to the function, as the compiler will also decay the array name to a pointer when used as a function argument: 同样,你可以通过
a
直的功能,作为函数参数一起使用时,编译器也将衰减数组名的指针:
int a[] = {1,2,3};
int *p = &(a[0]);
f(p, 3)
f(a, 3); // these two are equivalent
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