＆QUOT;＆的地址QUOT; （安培）的数组/地址被就是gcc忽略？ [英] "Address of" (&) an array / address of being ignored be gcc?

问题描述

我是一名编程课程的助教，有的学生做这种类型的错误：

I am a teaching assistant of a introductory programming course, and some students made this type of error:

char name[20];
scanf("%s",&name);

这是因为他们的学习也就不足为奇了......令人惊讶的是，除了海湾合作委员会警告后，code ++工程（至少这部分）。我一直在试图理解和我写了下面code：

which is not surprising as they are learning... What is surprising is that, besides gcc warning, the code works (at least this part). I have been trying to understand and I wrote the following code:

void foo(int *v1, int *v2) {
  if (v1 == v2)
    printf("Both pointers are the same\n");
  else
    printf("They are not the same\n");
}

int main() {
  int test[50];
  foo(&test, test);
  if (&test == test)
    printf("Both pointers are the same\n");
  else
    printf("They are not the same\n");
}

编译和执行：

$ gcc test.c -g
test.c: In function ‘main’:
test.c:12: warning: passing argument 1 of ‘foo’ from incompatible pointer type
test.c:13: warning: comparison of distinct pointer types lacks a cast
$ ./a.out 
Both pointers are the same
Both pointers are the same

任何人都可以解释为什么他们不一样？

Can anyone explain why they are not different?

我怀疑这是因为我不能让一个数组的地址（因为我不能让＆放大器;＆放大器; X ），但在这种情况下，code不应该编译。

I suspect it is because I cannot get the address of an array (as I cannot have & &x), but in this case the code should not compile.

编辑：我知道自身的数组是一样的第一个元素的地址，但这不是与此相关的问题，我想。例如：

I know that an array by itself is the same as the address of the first element, but this is not related to this problem, I think. For example:

int main() {
  int a[50];
  int * p = a;
  printf("%d %d %d\n", p == a, p == &a[0], &p[0] == a);
  printf("%d %d %d\n", p == &a, &p == a, &p == &a);
}

打印：

$ ./a.out 
1 1 1
1 0 0

我不明白，为什么第二行以 1 开始。

推荐答案

在您的例子中，数组测试是50 整数块。所以它看起来是这样的：

In your example, the array test is a block of 50 ints. So it looks like this:

| int | int | ... | int |

在应用一元＆安培; 运营商数组，你得到的阵列的地址的。当你把它应用到别的只是想，真的。因此，＆放大器;测试是指向50 整数该块的指针：

When you apply the unary & operator to an array, you get the address of the array. Just like when you apply it to anything else, really. So &test is a pointer that points to that block of 50 ints:

(&test) -----------> | int | int | ... | int |

指向50整数数组的类型为指针 INT（*）[50] - 这是＆放大器的类型;测试。

当你只需要使用名称测试在它不是操作数的任何地方无论是的sizeof 或一元 - ＆安培; 运营商，它是计算为一个指向它的第一个元素。因此，测试传递给富（）计算为一个指向试验[ 0] 元素：

When you just use the name test in any place where it's not the operand of either the sizeof or unary-& operators, it is evaluated to a pointer to its first element. So the test that you pass to foo() evaluates to a pointer to the test[0] element:

(test) -----------------\
                        v
(&test) -----------> | int | int | ... | int |

您可以看到，这些都指向同一个地址 - 尽管＆放大器;测试指向整个数组，而测试指向数组的第一个元素。

You can see that these both are pointing to the same address - although &test is pointing to the whole array, and test is pointing to the first element of the array (which only shows up in the different types that those values have).

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