sizeof()运算符和Poiner声明中的问题 [英] Problem in sizeof() operator and poiner declaration

问题描述

这个问题是在我的Sem-2考试中提出的.问题要求我们提供所需的输出.

This question was asked in my Sem-2 examination. Question asked us to give the desired output.

int main(void)
{
    int a[] = {10,20,30,40,50,60};

    int (*p1)[2]=a , (*p2)[3]= a;

    if(sizeof(p1)==sizeof(p2))
    printf("%d",*(*p1+2));

    if(sizeof(*p1)==sizeof(*p2))
    printf("%d",*(*(p2+1)));

    return(0);
}

编译器警告:

Warning: initialization from incompatible pointer type [-Wincompatible-pointer-types]
         initialization from incompatible pointer type [-Wincompatible-pointer-types]

我期望的输出:20

运行时得到的输出:30

Output that I get when I run it: 30

使用:gcc(Ubuntu 7.4.0-1ubuntu1〜18.04.1)7.4.0

Using : gcc (Ubuntu 7.4.0-1ubuntu1~18.04.1) 7.4.0

推荐答案

让我们忽略未定义的行为，找出可能发生的情况.

Let's ignore the undefined behavior to work out what is probably happening.

p1和p2都指向a[0](忽略不兼容的指针类型).

p1 and p2 are both pointing to a[0] (ignoring the incompatible pointer types).

p1和p2都是指针.指向对象类型的指针通常具有相同的大小(假设是这种情况)，因此sizeof(p1)==sizeof(p2)将为true.

p1 and p2 are both pointers. Pointers to object types are generally the same size (assume this is the case), so sizeof(p1)==sizeof(p2) will be true.

p1的类型为int (*)[2]，因此*p1的类型为int[2].在大多数表达式中，数组将衰减为其第一个元素的指针，因此在表达式*(*p1+2)中，*p1将衰减为int *并指向a[0].因此，*p1+2将指向a[2].因此，*(*p1+2)将与a[2]相同，其值为30.因此，程序将打印30.

p1 is of type int (*)[2], so *p1 is of type int[2]. In most expressions, an array will decay to a pointer to its first element, so in the expression *(*p1+2), *p1 will decay to an int * and will be pointing to a[0]. Therefore *p1+2 will be pointing to a[2]. Therefore *(*p1+2) will be the same as a[2], which has the value 30. Therefore the program prints 30.

当数组是sizeof运算符的操作数时，它不会衰减到指针. *p1的类型为int[2]，而*p2的类型为int[3]，因此sizeof(*p1)==sizeof(*p2)等效于sizeof(int[2])==sizeof(int[3])，这是错误的.因此，不会评估第二次打印*(*p2+1)值的printf调用.

An array does not decay to a pointer when it is the operand of the sizeof operator. *p1 is of type int[2] and *p2 is of type int[3], so sizeof(*p1)==sizeof(*p2) is equivalent to sizeof(int[2])==sizeof(int[3]), which is false. Therefore the second printf call that prints the value of *(*p2+1) is not evaluated.

(假设第二个printf被调用并且对*(*p2+1)进行求值.*p2的类型为int[3]，在该表达式中，它衰减为指向a[0]的int *.因此，指向a[1].因此，*(*p2+1)将与a[1],相同，其值为20.)

(Let's pretend the second printf is called and that *(*p2+1) is evaluated. *p2 is of type int[3] and in this expression it decays to an int * pointing to a[0]. Therefore *p2+1 points to a[1]. Therefore, *(*p2+1) will be the same as a[1], which has the value 20.)

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