假设我们有结构填充,int的大小为4,double的大小为8字节,下面这段代码中结构的大小是多少

Question

任何人都可以告诉我,下面显示的结构的大小是24而不是20.

typedef struct
{
    double d;  // this would be 8 bytes
    char c;   // This should be 4 bytes considering 3 bytes padding
    int a;   // This would be 4 bytes
    float b; // This would be 4 bytes
} abc_t;

main()
{
    abc_t temp;
    printf("The size of struct is %d\n",sizeof(temp));
}

我的假设是当我们考虑填充时结构的大小将是20但是当我运行此代码时,大小打印为24.

Answer 1

大小将是24.这是因为最后一个成员用所需的字节数填充,因此结构的总大小应该是任何结构成员的最大对齐的倍数.

所以填充就像

typedef struct
{
    double d;  // This would be 8 bytes
    char c;    // This should be 4 bytes considering 3 bytes padding
    int a;     // This would be 4 bytes
    float b;   // Last member of structure. Largest alignment is 8.  
               // This would be 8 bytes to make the size multiple of 8 
} abc_t;

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