用于打印数组中元素位置的 C 函数不起作用

Question

我试图打印两个 vector 的标量积、每个 vector 中最大元素的值和位置，以及每个 vector 中最小元素的值和位置。但是，我用于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，因为它使用与我用于查找最大值和最大值位置的函数相同的语法，并且它打印正确的数字。这就是我的代码的样子:

#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

但我的输出如下所示:

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”，即位置？

Answer 1

你的位置函数不起作用，因为你在搜索开始时没有将 pos 初始化为零，所以如果最小值在条目零中，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，您将整数与 double 进行比较，并且会得到错误的结果。

#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