明解C语言 入门篇 第七章答案

练习7-1

#include <stdio.h>

int main() {
    int n;
    printf("%d\t%d\t%d\n", sizeof ,sizeof(unsigned)-,sizeof n+ ); //此行显示结果为 4 3 6 因为1的字节就是为4，而-1的字节也是4再减去-1所以显示为3，最后是n+2为6
    printf("%d\t%d\t%d\n", sizeof +, sizeof(double) - , sizeof(n + ));//此行显示结果是 4 7 4 因为1的字节是4，double的字节长是8 -1是7，把（n+2）括起来之后使其为一个值，所以是4；
    printf("%d\t%d\t%d\n", sizeof - , sizeof((double)-), sizeof (n + 2.0)  );//此行结果是 4 8 8 因为1的字节是4，把double-1都括起来之后，就是double类型字节为8，同理最后也是8
}

练习7-2

#include <stdio.h>
#include<math.h>
int main() {
    unsigned int number;
    int x;
    int i;
    int number1;

    printf("请输入初始值:" );
    scanf("%d", &number);
    number1 = number;
    printf("左移多少位:");
    scanf("%d", &x);
    number <<= x;
    printf("左移后的值:%d",number);
    putchar('\n');
    for (i = ; i < x; i++) {
        number1 *= ;
    }
    printf("乘以2的指数幂的值:%d", number1);

}

#include <stdio.h>
#include<math.h>
int main() {
    unsigned int number;
    int x;
    int i;
    int number1;

    printf("请输入初始值:" );
    scanf("%d", &number);
    number1 = number;
    printf("右移多少位:");
    scanf("%d", &x);
    number >>= x;
    printf("右移后的值:%d",number);
    putchar('\n');
    for (i = ; i < x; i++) {
        number1 /= ;
    }
    printf("除以2的指数幂的值:%d", number1);

}

练习7-3

#include <stdio.h>
#include<math.h>
unsigned rrotate(unsigned x,int n){
    x >>=n;
    return x;
}

int main() {
    unsigned int number;
    int x;
    printf("请输入初始值:");
    scanf("%d", &number);
    printf("右移多少位:");
    scanf("%d", &x);

    printf("右移后的值:%d", rrotate(number, x));

}

#include <stdio.h>
#include<math.h>
unsigned lrotate(unsigned x,int n){
    x <<=n;
    return x;
}

int main() {
    unsigned int number;
    int x;
    printf("请输入初始值:");
    scanf("%d", &number);
    printf("左移多少位:");
    scanf("%d", &x);

    printf("左移后的值:%d", lrotate(number, x));

}

练习7-4

unsigned set(unsigned x, int pos)
{
    unsigned mask = (1u << pos);
    x = (mask | x);
    return x;
}
//将第pos位设置为1

unsigned reset(unsigned x, int pos) {
    unsigned mask = 1u;

    if (x & (1u<<pos))
        return (x ^ (1u << pos));
    else
        return x;

}
//将第pos为设置为0

unsigned inverse(unsigned x, int pos) {
    unsigned mask = 1u << pos;
    x = (x ^ mask);
    return x;

}
//将第pos位取反

练习7-5

unsigned set_n(unsigned x, int pos, int n) {

    int i;
    for (i = ; i < n - pos; i++) {
        unsigned mask = (1u << (pos + i));
        x = (mask | x);
    }
    return x;
}
//一个位段都设置为1

unsigned reset_n(unsigned x, int pos, int n) {
    unsigned mask = 1u;
    int i;
    for (i = ; i < n-pos; i++) {
        if (x & (1u<<pos+i))
            x= (x ^ (1u << pos+i));
    }
    return x;

}
//一个位段都设置为0

unsigned inverse_n(unsigned x, int pos, int n) {
    int i;
    unsigned mask = 1u;
    for (i = ; i < n - pos; i++) {
        x = (x ^ (mask << pos + i));
    }
    return x;
}
//一个位段都取反

练习7-6

#include <stdio.h>
// 显示unsigned类型的值最大值，同时验证最大值+1是0.
int main() {
    unsigned a = ;
    while (++a > );
        printf("%u", a - );
        putchar('\n');
        printf("%u", a);
}

练习7-7

#include <stdio.h>

int main() {
    float a;
    double b;
    long double c;

    scanf("%f",&a);
    printf("a=%f", a);
    putchar('\n');
    scanf("%lf", &b);
    printf("b=%f", b);
    putchar('\n');
    scanf("%lf", &c);
    printf("c=%f", c);

}

练习7-8

#include <stdio.h>

int main() {
    float a;
    double b;
    long double c;
    printf("%d,%d,%d",sizeof(a), sizeof(b), sizeof(c));

}

练习7-9

#include <stdio.h>
#include<math.h>

int main() {
    double m;
    double x;
    printf("正方形的面积是：");
    scanf("%lf", &m);
    printf("边长是：%.2f", sqrt(m));

}

练习7-10

#include <stdio.h>
int main(void)
{
    float x;
    int i;
    float m = 0.0;

    for (i = ; i <= ; i++) {
        x = i / 100.0;
        printf("x=%f x=%f\n",m,x);
        m += 0.01;
    }

    return ;
}

练习7-11

#include <stdio.h>
int main(void)
{
    float x;
    int i;
    float m = 0.0;
    float sum1 = ;
    float sum2 = ;

    for (i = ; i <= ; i++) {
        x = i / 100.0;
        printf("x=%f x=%f\n",m,x);
        m += 0.01;
        sum1 += m;
        sum2 += x;
    }
    printf("%f,%f", sum1, sum2);
    return ;
}