用于打印数组中元素位置的C函数不起作用答案

【问题标题】：C function for printing the position of an element in an array not working用于打印数组中元素位置的C函数不起作用
【发布时间】：2020-02-01 02:01:24
【问题描述】：

我正在尝试打印两个向量的标量积，即每个向量中最大元素的值和位置，以及每个向量中最小元素的值和位置。但是，我的查找最小值和最小值位置的函数不起作用，我不确定为什么，因为它使用与查找最大值和最大值位置的函数相同的语法，并且打印正确数字。这是我的代码的样子：

#include <stdio.h>
#include <stdlib.h>

double findingmax(double *arr, int n){
    int max = arr[0];
    for(int i = 0; i < n; i++){
        if(arr[i] > max){
            max = arr[i];
        }
    }
    return max;
}

int findingmaxpos(double *arr, int n){
    int max = arr[0];
    int pos;
    for(int i = 0; i < n; i++){
        if(arr[i] > max){
            max = arr[i];
            pos = i;
        } 
    }
    return pos;
} 

double findingmin(double *arr, int n){
    int min = arr[0];
    for(int i = 0; i < n; i++){
        if(arr[i] < min){
            min = arr[i];
        }
    }
    return min;
}

int findingminpos(double *arr, int n){
    int min = arr[0];
    int pos;
    for(int i = 0; i < n; i++){
        if(arr[i] < min){
            min = arr[i];
            pos = i; 
        }
    }
    return pos; 
} 

double scalarproduct(double *v, double *w, int n){
    double vw[n];
    for(int i = 0; i < n; i++){
        vw[i] = (v[i] * w[i]); 
    }
    double scalprod = 0; 
    for(int i = 0; i < n; i++){
        scalprod += vw[i];
    }
    return scalprod;
}
int main(){
    int n;
    scanf("%d", &n);
    double *v; 
    v = (double *) malloc(sizeof(double) * n);
    double *w;
    w = (double *) malloc(sizeof(double) * n);
    for(int i = 0; i < n; i++){
        scanf("%lf", &v[i]);       
    }
    for (int i = 0; i < n; i++){
        scanf("%lf", &w[i]);
    }
    printf("Scalar product=%lf\n", scalarproduct(v, w, n));
    printf("The smallest = %lf\n", findingmin(v, n));
    printf("Position of the smallest = %d\n", findingminpos(v, n));
    printf("The largest = %lf\n", findingmax(v, n));
    printf("Position of the largest = %d\n", findingmaxpos(v, n));
    printf("The smallest = %lf\n", findingmin(w, n));
    printf("Position of the smallest = %d\n", findingminpos(w, n));
    printf("The largest = %lf\n", findingmax(w, n));
    printf("Position of the largest = %d\n", findingmaxpos(w, n));
    return 0; 
}

输入是这样的：

3
1.1
2.5
3.0
1.0
1.0
1.0

输出应该是这样的：

Scalar product=6.600000
The smallest = 1.100000
Position of smallest = 0
The largest = 3.000000
Position of largest = 2
The smallest = 1.000000
Position of smallest = 0
The largest = 1.000000
Position of largest = 0

但我的输出是这样的：

Scalar product=6.600000
The smallest = 1.000000
Position of the smallest = 32766
The largest = 3.000000
Position of the largest = 2
The smallest = 1.000000
Position of the smallest = 32766
The largest = 1.000000
Position of the largest = 32766

如何打印正确的“i”，位置？

【问题讨论】：

您的位置函数不起作用，因为您没有在搜索开始时将 pos 初始化为零，因此如果最小的在条目零中，则 pos 未初始化（并设置为堆栈如 32766)
另外，你的局部变量 max 和 min 必须声明为 double，否则你是在比较 ints 和 doubles，并返回一个向下取整的值。

标签： c arrays memory dynamic allocation

【解决方案1】：

您的位置函数不起作用，因为您没有在搜索开始时将 pos 初始化为零，因此如果最小的在条目零中，则 pos 未初始化（并设置为堆栈上的任何内容，例如 32766.初始化 pos=0 就可以了。

此外，您需要将局部变量max 和min 更改为double，而不是int。否则，您将 int 与 double 进行比较，您会得到错误的结果。

#include <stdio.h>
#include <stdlib.h>

double findingmax(double *arr, int n){
    double max = arr[0];
    for(int i = 0; i < n; i++){
        if(arr[i] > max){
            max = arr[i];
        }
    }
    return max;
}

int findingmaxpos(double *arr, int n){
    double max = arr[0];
    int pos = 0;
    for(int i = 0; i < n; i++){
        if(arr[i] > max){
            max = arr[i];
            pos = i;
        } 
    }
    return pos;
} 

double findingmin(double *arr, int n){
    double min = arr[0];
    for(int i = 0; i < n; i++){
        if(arr[i] < min){
            min = arr[i];
        }
    }
    return min;
}

int findingminpos(double *arr, int n){
    double min = arr[0];
    int pos = 0;
    for(int i = 0; i < n; i++){
        if(arr[i] < min){
            min = arr[i];
            pos = i; 
        }
    }
    return pos; 
} 

double scalarproduct(double *v, double *w, int n){
    double vw[n];
    for(int i = 0; i < n; i++){
        vw[i] = (v[i] * w[i]); 
    }
    double scalprod = 0; 
    for(int i = 0; i < n; i++){
        scalprod += vw[i];
    }
    return scalprod;
}
int main(){
    int n;
    scanf("%d", &n);
    double *v; 
    v = (double *) malloc(sizeof(double) * n);
    double *w;
    w = (double *) malloc(sizeof(double) * n);
    for(int i = 0; i < n; i++){
        scanf("%lf", &v[i]);       
    }
    for (int i = 0; i < n; i++){
        scanf("%lf", &w[i]);
    }
    printf("Scalar product=%lf\n", scalarproduct(v, w, n));
    printf("The smallest = %lf\n", findingmin(v, n));
    printf("Position of the smallest = %d\n", findingminpos(v, n));
    printf("The largest = %lf\n", findingmax(v, n));
    printf("Position of the largest = %d\n", findingmaxpos(v, n));
    printf("The smallest = %lf\n", findingmin(w, n));
    printf("Position of the smallest = %d\n", findingminpos(w, n));
    printf("The largest = %lf\n", findingmax(w, n));
    printf("Position of the largest = %d\n", findingmaxpos(w, n));
    return 0; 
}

产生输出：

Scalar product=6.600000
The smallest = 1.100000
Position of the smallest = 0
The largest = 3.000000
Position of the largest = 2
The smallest = 1.000000
Position of the smallest = 0
The largest = 1.000000
Position of the largest = 0

【讨论】：