C第十章（数组和指针）习题

末蓝、 2023-07-04 12:51 133阅读 0赞

1.  int torf\[2\]\[2\] = \{12, 14, 16\};是什么意思  
    我竟然不确定？？？我个垃圾  
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#include <stdio.h>
    int main()
    { 
        int *ptr;
        int torf[2][2] = { 12, 14, 16};
        ptr = torf[0];//指向torf[0][0]
        return 0;
    }

![在这里插入图片描述][20200211121419664.png]  
就算只写一个元素也要括起来，否则出错

int torf[2][2] = { 12, { 14, 16}};

![在这里插入图片描述][20200211121728438.png]

int torf[2][2] = { { 12}, { 14, 16}};

![在这里插入图片描述][20200211121747307.png]

#include <stdio.h>
    int main()
    { 
        int *ptr;
        int torf[2][2] = { { 12}, { 14, 16}};
        ptr = torf[0];
        printf("*ptr=%d, *(ptr+2)=%d\n", *ptr, *(ptr+2));
        return 0;
    }

*ptr=12, *(ptr+2)=14

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1,2不重要，重要的是3中易错，ref+1指向4，但是++ref不是有效表达式！！！不可以改变数组名的地址，人家是常量，改不了

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不明白？？？？？？为啥取址就是一种表示？？我以为题目是让我们用指针表示法表示  
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我的答案

#include <stdio.h>
    int main()
    { 
        int torf[3][2] = { { 12}, { 14, 16},{ 2, 6}};
        printf("torf[2][1] = %d, *(*(torf+2)+1) = %d\n", torf[2][1], *(*(torf+2)+1));
        return 0;
    }

torf[2][1] = 6, *(*(torf+2)+1) = 6

a. *(*(grid+22)+56)//grid是指向第一个子数组，grid+22指向第23个子数组，即grid[22],
    *(grid+22)等于grid[22],grid[22]是一个内含100个元素的数组的首元素地址，grid[22]+56则是指向第57个元素
    
    b1.*grid[22]//grid[22]就是第23个数的首元素grid[22][0]的地址
    b2. *(grid+22)
    
    c1.**grid
    c2.*grid[0]
    c3.can't think of anything····

# \[\]的优先级高于\* #

声明下面的变量  
![在这里插入图片描述][20200211130548737.png]

int *psa[20];//相当于int *(psa[20]), []优先级高于*，所以psa先和[20]结合，说明psa是个数组
    
    char (*pstr)[20];//它是指针，所以圆括号括起来使得先和*结合

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int a[100] = { [99]=-1};
    int b[100] = { [5]=101, [10]=101, [11]=101, [12]=101, [3]=101};
    int b[100] = { [5]=101, [10]=101, 101, 101, [3]=101};//简写方式

#include <stdio.h>
    int main()
    { 
        int a[100] = { [99]=-1};
        int b[100] = { [5]=101, [10]=101, 101, 101, [3]=101};
        printf("a[99]=%d,b[5]=%d, b[10]=%d, b[11]=%d, b[12]=%d, b[3]=%d\n",
                a[99], b[5], b[10], b[11], b[12], b[3]);
        return 0;
    }

a[99]=-1,b[5]=101, b[10]=101, b[11]=101, b[12]=101, b[3]=101

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答案认为第二个是无效，我觉得答案错了，验证发现确实有效

#include <stdio.h>
    int main()
    { 
        int a[100];
        scanf("%d", &a);
        printf("a[0]=%d\n", a[0]);
        return 0;
    }

22
    a[0]=22

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show((double [4]) { 8, 3, 9, 2}, 4);
    show2((double [][3]) { { 8, 3, 9},{ 5, 4, 1}}, 2);

![在这里插入图片描述][20200211133139547.png]  
这次改用指针表示法重写还发现了之前的一个错误，之前写的\*(\*(rainfall+i)+j)!!!不对哦

而且之前求月均还除的是月！！应该除以年

#include <stdio.h>
    #define YEAR 3
    #define MONTH 4
    int main()
    { 
        float sum_year[YEAR] = { }, sum_month[MONTH]={ };
        float total=0;
        float rainfall[YEAR][MONTH] = { { 3.4, 4.5, 4.2, 5.2},{ 4.6, 6.4, 7.3, 4.8}, { 2.1, 3.1, 8.2, 5.2}};
        printf("Year Rainfall(inches)\n");
        for(int i=0;i<YEAR;i++)
        { 
            for(int j=0;j<MONTH;j++)
                *(sum_year+i) += *(*(rainfall+i)+j);
    
            printf("%d: %.2f\n", 2011+i, *(sum_year+i));
            total += *(sum_year+i);
        }
        printf("The yearly average is %.2f inches.\n\n", total/YEAR);
        printf("Monthly averages:\n");
        printf("Jan Feb Mar Apr\n");
        for(int i=0;i<MONTH;i++)
        { 
            for(int j=0;j<YEAR;j++)
               *(sum_month+i) += *(*(rainfall+j)+i);//这次改用指针表示法重写还发现了之前的一个错误，之前写的*(*(rainfall+i)+j)!!!不对哦
            printf("%.2f ", *(sum_month+i)/YEAR);//之前除的MONTH，不对哦
        }
        return 0;
    }

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2题， 2-5的代码都是我重写了一遍的，刚才浏览器突然卡了，然后就啥都没了，气哭了，现在又耐着性子重写，不过我发现重写一遍也好，我发现我还是出了两次错，就算巩固了

对于原数组，我用了const保护，注意了搭配同款const指针，0警告

听说华为编程规范要求就算if, for, while只有一个语句也要用花括号，所以就用了

听说规范不允许在while条件里写n–>0这种略显复杂的表达式，要老老实实写在花括号里面，所以也就听话了，向规范低头

#include <stdio.h>
    void copy_arr(double t[], const double s[], int n);
    void copy_ptr(double *t, const double *s, int n);
    void copy_ptrs(double *t, const double *s, double *tail);
    void show_arr(const double *arr, int n);
    int main()
    { 
        double source[5] = { 1.1, 2.2, 3.3, 4.4, 5.5};
        double target1[5], target2[5], target3[5];
        copy_arr(target1, source, 5);
        copy_ptr(target2, source, 5);
        copy_ptrs(target3, source, source+5);
        printf("source: ");
        show_arr(source, 5);
        printf("target1: ");
        show_arr(target1, 5);
        printf("target2: ");
        show_arr(target2, 5);
        printf("target3: ");
        show_arr(target3, 5);
        return 0;
    }
    
    void copy_arr(double t[], const double s[], int n)
    { 
        int i;
        for(i=0;i<n;i++)
        { 
            t[i] = s[i];
        }
    }
    
    void copy_ptr(double *t, const double *s, int n)
    { 
        double *ptr_t = t;
        const double *ptr_s = s;
        int i;
        for(i=0;i<n;i++)
        { 
            *ptr_t++ = *ptr_s++;
        }
    }
    
    void copy_ptrs(double *t, const double *s, double * tail)
    { 
        int d = tail - s;//指针减法，得到数组元素个数
        double *ptr_t = t;
        const double *ptr_s =  s;
        while(d>0)
        { 
            *ptr_t++ = *ptr_s++;
            d--;
        }
    }
    
    void show_arr(const double *arr, int n)
    { 
        int i;
        for(i=0;i<n;i++)
        { 
            printf("%.2f ", arr[i]);
        }
        putchar('\n');
    }

source:  1.10 2.20 3.30 4.40 5.50
    target1: 1.10 2.20 3.30 4.40 5.50
    target2: 1.10 2.20 3.30 4.40 5.50
    target3: 1.10 2.20 3.30 4.40 5.50

3题

#include <stdio.h>
    void arr_max(const int *arr, int n);
    int main()
    { 
        int arr[] = { 1, 12, 73, 54, 5};
        arr_max(arr, 5);
        return 0;
    }
     void arr_max(const int *arr, int n)
     { 
         int m = *arr;
         const int *ptr = arr;
         while(--n>0 && *ptr>=m)
         { 
             m = *ptr++;
         }
         printf("maximum: %d", m);
     }

maximum: 73

4题

#include <stdio.h>
    void arr_max(const double *arr, int n);
    int main()
    { 
        double arr[] = { 1.1, 10.3, 3.6, 7.3, 5.9};
        arr_max(arr, 5);
        return 0;
    }
     void arr_max(const double *arr, int n)
     { 
         double m = *arr;
         int sub=0, n1=n;
         const double *ptr = arr;
         while(--n>0 && *ptr>=m)
         { 
             m = *ptr++;
             sub = n1-n-1;
         }
         printf("maximum: %.1f\nsubscript: %d\n", m, sub);
     }

maximum: 10.3
    subscript: 1

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5题

#include <stdio.h>
    void diff(const double *arr, int n);
    int main()
    { 
        const double arr[]={ 3.1, 78.2, 9.2, 7.1, 0.9};
        diff(arr, sizeof arr / sizeof arr[0]);
        return 0;
    }
    
    void diff(const double * arr, int n)
    { 
        double max, min, d;
        int i;
        max = min = *arr;//多重赋值
        const double *ptr = arr;
        for(i=0;i<n;i++)
        { 
            if(*(ptr+i)>max)
                max = *(ptr+i);
            else if(arr[i]<min)
                min = arr[i];
    
        }
        d = max - min;
        printf("maximum:%.2f\nminimum:%.2f\ndifference: %.2f\n", max, min, d);
    }

maximum:78.20
    minimum:0.90
    difference: 77.30

#include <stdio.h>
    void reverse(double *ar, int n);
    int main()
    { 
        double lazy[] = { 10.2, 3.4, 6.3, 76.1, 0.6};
        reverse(lazy, 5);
        return 0;
    }
    
    void reverse(double *ar, int n)
    { 
        double *ptr=ar+n;
        while(n>0)
        { 
            printf("%.1f ", *--ptr);
            n--;
        }
    }

0.6 76.1 6.3 3.4 10.2

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#include <stdio.h>
    #define SIZE 4
    void add_arr(int *ar1, int *ar2, int *ar3, int n);
    int main()
    { 
        int apple[SIZE] = { 1, 2, 3, 4};
        int pear[SIZE] = { 4, 3, 2, 1};
        int sum[SIZE] = { };
        add_arr(apple, pear, sum, SIZE);
        return 0;
    }
    
    void add_arr(int *ar1, int *ar2, int *ar3, int n)
    { 
        int i;
        int *ptr3=ar3, *ptr1=ar1, *ptr2=ar2;
        for(i=0;i<n;i++)
        { 
            *(ptr3+i) = *(ptr1+i) + *(ptr2+i);
            printf("%d ", *(ptr3+i));
        }
    }

5 5 5 5

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