Question

error C2664: 'f' : cannot convert parameter 1 from 'int [5]' to 'int'This conversion

requires a reinterpret_cast, a C-style cast or function-style cast

#include<stdio.h>
int main()
{void f(int m,int n);
int i;
int a[5],b[5],c[5];
printf("请输入数组a：\n");
for(i=0;i<5;i++)
{printf("a%d=",i+1);
scanf("%d",&a[i]);}
printf("请输入数组b：\n");
for(i=0;i<5;i++)
{printf("b%d=",i+1);
scanf("%d",&b[i]);}
printf("两数组之积c:\n");
f(a,b);
return 0;}

void f(int m[],int n[])
{int i,c[5];

for(i=0;i<5;i++)
{c[i]=m[i]*n[i];
printf("c%d=%d\t",i+1,c[i]);}
}

Answer 1

#include<stdio.h>

int main()
{ void f(int m[],int n[]);//这里写错了，看看好，给你改过来，现在可以了
int i;
int a[5],b[5],c[5];
printf("请输入数组a：\n");
for(i=0;i<5;i++)
{printf("a%d=",i+1);
scanf("%d",&a[i]);}
printf("请输入数组b：\n");
for(i=0;i<5;i++)
{printf("b%d=",i+1);
scanf("%d",&b[i]);}
printf("两数组之积c:\n");
f(a,b);
return 0;
}

void f(int m[],int n[])
{int i,c[5];

for(i=0;i<5;i++)
{
c[i]=m[i]*n[i];
printf("c%d=%d\t",i+1,c[i]);}
}