使用函数并从数组中检查更大的值

Question

我试图使用数组创建一个函数并获得更大的价值，但是我却失败了并获得这些消息，有人可以帮助我吗？

\\ Untitled4.c在函数“ main”中：

25 27 \\ Untitled4.c [警告]传递'quatroMaior'的参数1使指针从整数开始而没有强制转换

3 5 \\ Untitled4.c [注]预期为'int *'，但参数的类型为'int'

#include <stdio.h>

int greater(int array[]) {
    int i, greater;

    for (i = 0; i < 4; i++) {
        if(array[i] > greater) {
            greater = array[i];
        }
    }

    return greater;
}

int main(void) {

    int j, v[4];

    printf("Type four values\n");

    for (j = 0; j < 4; j++) {
        scanf("%d", &v[j]);
    }

    printf("%d", greater(v[4]));

    system("pause");
    return 0;
}

Answer 1

正如约瑟夫在评论中所说，您传递的是整数，而不是数组。 您必须编写greater(v)而不是greater(v[4]) 。

接下来需要做的是将greater变量初始化为greater = 0 。

#include <stdio.h>

int greater(int array[])
{
    int i, greater = 0;

    for (i = 0; i < 4; i++)
    {
        if (array[i] > greater)
        {
            greater = array[i];
        }
    }

    return greater;
}

int main (void)
{
    int j, v[4];

    printf("Type four values\n");
    for (j = 0; j < 4; j++)
    {
        scanf("%d", &v[j]);
    }

    printf("%d\n", greater(v));

    return 0;
}

Answer 2

#include <stdio.h>

int greater(int array[]) {
    int i, greater;
/* greater is not init so you have an Undefined behavior.
** You could init it with array[0] but remind that need that size of array is at least 1.*/

    for (i = 0; i < 4; i++) { // If you do start at 1 because 0 is already in greater.
        if (array[i] > greater) { // here greater is not init !
            greater = array[i]; 
        }
    }

    return greater;
}

int main(void) {

    int j, v[4];

    printf("Type four values\n");

    for (j = 0; j < 4; j++) {
        scanf("%d", &v[j]);
    }

    printf("%d", greater(v[4]));
/* you send v[4] but you array has only range from 0 to 3.
** And is not what greater function expect he expect a array so just v.*/

    system("pause");
    return 0;
}