小编mra*_*hok的帖子

当我在结构中定义字符类型时,它似乎需要超过1个字节; 实际上它似乎需要4个字节.

以下是我的计划:

#include <stdio.h>

int main(void)
{
    struct book{
        char name;
        float price;
        int pages;
    };
    struct book b1={'B',130.00,550};
    printf("\nAddress of structure:%u",&b1);
    printf("\nAddress of character name:%u",&b1.name);
    printf("\nAddress of float price:%u",&b1.price);
    printf("\nAddress of integer pages:%u",&b1.pages);
    printf("\n\n");
    return 0;
}

当我运行上面的程序时,我得到以下输出:

    Address of structure:557762432
    Address of character name:557762432
    Address of float price:557762436
    Address of integer pages:557762440

为什么我看到变量"name"的地址和变量"price"之间的4个字节的差异？

运行该程序的系统是运行Fedora-14的x86_64位.

我想使用如下字典:示例:{[8,16]:[[1,2,4,8],8],[16,24]:[[1,2,3,4,8, 12],12]}

8和16是将要输入的两个数字,我需要构建如上所述的字典.

使用setdefault,我可以在字典中创建值列表,但不能为键创建列表

以下是我的代码:

#!/usr/bin/env python
"""
        This Program calculates common factors between two Numbers , which 
        is stored on a list and also greatest common factor is also computed.
        All this is stored in a dictionary
        Example: { '[n1, n2]': [[Commonfac1(n1,n2), Commonfac2(n1,n2)....Commonfacn(n1,n2)],GreatestCommonFactor] }
"""
def Factors(number):
        result = []
        for i in range(1, number+1):
                if (number % i) == 0:
                        result.append(i)

        return result

def Common_Factors(n1, n2):
        result = []
        for element in n1:
                if element in n2:
                        result.append(element) …