Question

以下代码的输出是什么，我无法理解* x和x之间有什么区别，我认为是，是吗？

int a = 5 ;
int * xxx  ;
xxx = a;
printf("\n\n%d" , *xxx);
printf("\n\n%d" , xxx);
*xxx = a;
printf("\n\n%d" , *xxx);
printf("\n\n%d" , xxx);

Answer 1

您没有将a的地址分配给指针xxx，而是传递其值。此代码将调用未定义的行为。你会得到任何东西。改变

xxx = a;

到

x = &a;

现在x是指向a的指针，当一元运算符*与指针一起使用时，它会检索存储在地址x处的值。这称为解除引用。
还要使用%p说明符打印地址。

printf("\n\n%p" , (void *)x);

Answer 2

*x表示存储在x. x所占据的地址的引用值表示您引用x.所持有的地址来到你的代码，

int a = 5 ; // a contains value 5
int * xxx  ;// xxx is a integer pointer
xxx = a; //  you are assigning 5 to xxx. Leads to undefined behaviour
printf("\n\n%d" , *xxx); //here *x means *xxx? You are printing value stored at address 5 Which leads to undefined behaviour. 
printf("\n\n%d" , xxx); // It will print value 5
*xxx = a; // Here you are assigning 5 to address held by xxx which is 5. Undefined behaviour
printf("\n\n%d" , *xxx); //Printing value stored at the address held by xxx,Leads to undefined behaviour
printf("\n\n%d" , xxx); // printing address held by xxx

使用指针并用C语言赋值

2 个答案: