鑄造指針和它的地址爲整型指針

是什麼以下兩個任務之間的區別？鑄造指針和它的地址爲整型指針

int main() 
{ 
    int a=10; 
    int* p= &a; 

    int* q = (int*)p; <------------------------- 
    int* r = (int*)&p; <------------------------- 
}

我對這兩個聲明的行爲非常困惑。
什麼時候應該使用一個比其他？

來源

2012-07-12 Green goblin

int* q = (int*)p;

是正確的，雖然過於冗長。 int* q = p就足夠了。 q和p都是int指針。

int* r = (int*)&p;

是不正確（從邏輯上講，儘管它可能編譯），因爲&p是int**但r是int*。我想不出你想要這樣的情況。

來源

2012-07-12 11:33:34

int main() 
{ 
    int a=10; 
    int* p= &a; 

    int* q = p; /* q and p both point to a */ 
    int* r = (int*)&p; /* this is not correct: */ 
    int **r = &p; /* this is correct, r points to p, p points to a */ 

    *r = 0; /* now r still points to p, but p points to NULL, a is still 10 */ 
}

來源

2012-07-12 11:35:33 perreal

#include <stdio.h> 
int main() 
{ 
    int a = 10; /* a has been initialized with value 10*/ 

    int * p = &a; /* a address has been given to variable p which is a integer type pointer 
        * which means, p will be pointing to the value on address of a*/ 

    int * q = p ; /*q is a pointer to an integer, q which is having the value contained by p,      * q--> p --> &a; these will be *(pointer) to value of a which is 10; 

    int * r = (int*) &p;/* this is correct because r keeping address of p, 
         * which means p value will be pointer by r but if u want 
         * to reference a, its not so correct. 
         * int ** r = &p; 
         * r-->(&p)--->*(&p)-->**(&p)        
         */ 
     return 0; 
}

來源

2012-07-12 11:52:48 Ashutosh

類型無所謂。

表達p具有類型int *（指針int），因此表達&p具有類型int **（指針的指針的int）。這些不同，不兼容類型;您無法將類型int **的值分配給int *類型的變量，而無需進行明確的轉換。

的適當的事情是寫

int *q = p; 
int **r = &p;

爲什麼您需要將值轉換爲不同的類型，你不應該使用有明確的轉換在分配中，除非你知道。

來源

2012-07-12 14:31:53

鑄造指針和它的地址爲整型指針

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