Question

error LNK1120: 1 个无法解析的外部命令 错误 1 error LNK2019: 无法解析的外部符号

#include <stdio.h>

int input(float *array_number, unsigned n);
int average(float *array_number, unsigned n, float *aver);
int maxmin(float *array_number, unsigned n, float *max, float *min);
int main(void)
{
float date[10];
float aver = 0.0f, max = 0.0f, min = 0.0f;

input(date, 10);
average(date, 10, &aver);
maxmin(date, 10, &max, &min);

printf("aver=%f", aver);
printf("max=%f,min=%f", max, min);

return 0;
}

int input(float array_number, unsigned n)
{
printf("输入这个数组【10】：\n");
for (unsigned i = 0; i < n; ++i)
{
scanf_s("%d", array_number, 1);
}

return 0;
}
int average(float *array_number, unsigned n, float*aver)
{
float sum = 0.0f;
for (unsigned i = 0; i < n; ++i)
{
sum += *array_number;
}
*aver = (float)sum / 10;

return 0;
}
int maxmin(float *array_number, unsigned n, float *max, float *min)
{
for (unsigned i = 0; i<n; ++i)
{
if (*max<*(array_number + i))
{
*max = *(array_number + i);
}
if (*min>*(array_number + i))
{
*min = *(array_number + i);
}
}

return 0;
}
这个程序哪部分错了，跪求！！！！！！！

Answer 1

重新给你调整了一下，运行通过

#include <stdio.h>
int maxmin(float *array_number, unsigned n, float *max, float *min)
{
unsigned i = 0;
for (i = 0; i<n; ++i)
{
if (*max<*(array_number + i))
{
*max = *(array_number + i);
}
if (*min>*(array_number + i))
{
*min = *(array_number + i);
}
}

return 0;
}

int input(float* array_number, unsigned n)
{
printf("输入这个数组【10】：\n");
unsigned i = 0;
for (i = 0; i < n; ++i)
{
int d;
scanf("%d", &d);
*(array_number+i) = d;
}

return 0;
}
int average(float *array_number, unsigned n, float*aver)
{
float sum = 0.0f;
unsigned i = 0;
for (i = 0; i < n; ++i)
{
sum += *(array_number+i);
}
*aver = (float)sum / 10;

return 0;
}

int main(void)
{
float date[10] = {01};
float aver = 0.0f, max = -100000.0f, min = 100000.0f;

input(date, 10);
average(date, 10, &aver);
maxmin(date, 10, &max, &min);

printf("aver=%f", aver);
printf("max=%f,min=%f", max, min);

return 0;
}

追问

能给个理由吗，为甚吗这么做的话，link错误消除了
还有你可能没有认真看程序的功能，这么做很多问题

追答

嗯。。。因为你原来的程序报错太多，没法一一分辨是为什么，于是就把你主要的几个函数提了出来，再一一调试通过
实测录入10个数字，结果也是对的（主要是你一开始设定max=0，min=0，这样的话如果我不输入负数，min就永远是0；而如果我只输入负数，max就永远是0，因此把他们都先设了一个很大的正负数）