整型数据 long long int ，unsigned int 输出异常，求解！

Question

我对比了一下int，long，long long，unsigned 这个四种整型数据的不同的输出情况，有一下几个疑问：

1. unsigned int d1 = pow(2,32) 计算结果为 - 1。个人思考是：数据溢出，因为int只有最多能表达的数是4294967295，2的32次方大于4294967295所以直接取值4294967295，输出的是%d为有符号整型，故值为 - 1.
2. long long int 中 c1，c5，c6 的值，%ld 输出为什么是 0 和 - 1？long long 型字节数为8，即有64为存储空间，但是输出的数据却不大于2的32次方？实际上该如何输出？

以下为代码（新手上路，请海涵！使用编译器code blocks）：

#include <stdio.h>

int main()

{
    int a1, a2, a3, a4,a5,a6;
    a1 = pow(2,32);
    a2 = pow(2,32) - 2;
    a3 = pow(2,31);
    a4 = pow(2,31) - 2;

    long int b1, b2, b3, b4;
    b1 = pow(2,32);
    b2 = pow(2,32) - 2;
    b3 = pow(2,31);
    b4 = pow(2,31) - 2;

    long long int c1, c2, c3, c4, c5, c6;
    c1 = pow(2,32);
    c2 = pow(2,32) - 2;
    c3 = pow(2,31);
    c4 = pow(2,31) - 2;
    c5 = pow(2,43);
    c6 = pow(2,43)-1;

    unsigned int d1, d2, d3, d4;
    d1 = pow(2,32);
    d2 = pow(2,32) - 2;
    d3 = pow(2,31);
    d4 = pow(2,31) - 2;

    printf("  a1 = %d\n, a2 = %d\n, a3 = %d\n, a4 = %d\n",a1, a2, a3, a4);
    printf("\n  b1 = %d\n, b2 = %d\n, b3 = %d\n, b4 = %d\n",b1, b2, b3, b4);
    printf("\n  c1 = %d\n, c2 = %d\n, c3 = %d\n, c4 = %d\n, c5 = %d\n, c6 = %d\n",c1, c2, c3, c4, c5, c6);
    printf("\n  d1 = %d\n, d2 = %d\n, d3 = %d\n, d4 = %d\n",d1, d2, d3, d4);

    printf("\n  uc1 = %u\n, uc2 = %u\n, uc3 = %u\n, uc4 = %u\n, uc5 = %u\n",c1, c2, c3, c4, c5);
    printf("\n  ud1 = %u\n, ud2 = %u\n, ud3 = %u\n, ud4 = %u\n",d1, d2, d3, d4);

    printf("\n  ld1 = %ld\n, ld2 = %ld\n, lc1 = %ld\n, lc2 = %ld\n, lc5 = %ld\n, lc6 = %ld\n",d1, d2, c1, c2, c5);

    return 0;

}

以下为输出结果：

Answer 1

1. pow运算参数和返回值均为double类型，你赋值给int，进行了一次转换。
   参考：https://www.cplusplus.com/reference/cmath/pow/?kw=pow
   参考：https://blog.csdn.net/lin200753/article/details/27952897
2. printf输出会根据格式化的不同再次进行转换。例如%d则转为int, %u则转为unsigned int
3. 输出long long int，使用%lld
   参考：https://www.cplusplus.com/reference/cstdio/printf/?kw=printf