所以,到目前为止我有这个。我正在尝试找到数组中最大的两个数字并返回它们。我在网上查找了很多资源,其中大多数都说"参考电话"是要走的路。但是我不知道如何让它与我的程序一起工作。例如,我在网上看到了这个例子:
void Calculate(int x, int y, int* prod, int* quot)
{
*prod = x*y;
*quot = x/y;
}
int x = 10,y = 2, prod, quot;
Calculate(x, y, &prod, ")
上述程序实际上是如何"返回"的?如何将返回值打印到控制台?
#include "stdio.h"
void largest_two( int numbers[], int len, int *largest, int *next_largest){
int i, temp;
*largest = numbers[0];
*next_largest = numbers[1];
if(*largest < *next_largest){
temp = *next_largest;
*largest = *next_largest;
*next_largest = temp;
}
for (i=0; i<sizeof(numbers); i++) {
if(numbers[i]>= *largest){
*largest = numbers[i];
*next_largest = *largest;
}
else if ( numbers[i] > *next_largest){
*next_largest = numbers[i];
}
}
}
int main() {
int numbers[] = {3, 1, 2, 3, 6, 2, 8, 0, 0, 0};
int len = 3;
int largest, next_largest;
//==>??? printf("%d %d", largest_two(numbers, len, &largest, &next_largest));
}
Sides' 从指针问题(您应该阅读有关它们的教程/书籍(,您的主要问题是您正在尝试打印返回类型为 void
的函数的单个返回值,这意味着它根本不会返回。
您的代码:
int main() {
int numbers[] = {3, 1, 2, 3, 6, 2, 8, 0, 0, 0};
int len = 10; // sizeof(numbers)
int largest, next_largest;
largest_two(numbers, len, &largest, &next_largest);
printf("%d %d", largest, next_largest);
}
请记住,这仍然不完全正确,但它确实解决了打印数字的问题。
此外,传递len
意味着您不应该这样做for (i=0; i<sizeof(numbers); i++)
而是for (i=0; i<len; i++)
首先,这一行:
for (i=0; i<sizeof(numbers); i++)
不正确。您希望改为:
for (i=0; i<len; i++)
它应该作为sizeof numbers/sizeof numbers[0]
传递给largest_two()
,这是数组的实际长度。
我还建议将largest
和next_largest
设置为从<limits.h>
INT_MIN
,然后从他们的值中找到这些值。似乎您在使用指针时也遇到了问题,最好仅在需要时使用它们。
下面是一个简化方法的示例,该方法在数组的一个循环中查找最大和第二大元素。它也只在需要时使用指针。
法典:
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define ARRAYSIZE(x) (sizeof x/sizeof x[0])
void largest_two(int numbers[], size_t len, int *largest, int *next_largest);
int main(void) {
int numbers[] = {3, 1, 2, 3, 6, 2, 8, 0, 0, 0};
int largest, next_largest;
largest_two(numbers, ARRAYSIZE(numbers), &largest, &next_largest);
printf("largest = %dnnext_largest = %dn", largest, next_largest);
}
void largest_two(int numbers[], size_t len, int *largest, int *next_largest) {
int max, smax;
max = smax = INT_MIN;
for (size_t i = 0; i < len; i++) {
if (numbers[i] > max) {
smax = max;
max = numbers[i];
} else if (numbers[i] > smax && numbers[i] < max) {
smax = numbers[i];
}
}
*largest = max;
*next_largest = smax;
}
输出:
largest = 8
next_largest = 6
第二个数据集:
int numbers[] = {3, 1, 6, 3, 6, 2, 8, 0, 8, 7};
输出:
largest = 8
next_largest = 7