t=1-1/2-1/3-......-1/m
#include <stdio.h>
double fun (int m);
int main()
{
int m;
double result;
printf("m:\n");
scanf("%d",&m);
result=fun(m);
printf("The result is %lf\n",result);
return 0;
}
double fun(int m)
{int sum=0,i;
double t;
if(m>1){
for(i=2;i<=m;i++){
sum=sum+1/(double)i;
} t=1-sum;}
else t=1;
return t;
}

准确的来说是sum=sum+1/(double)i;这句语句根本就没运行,sum永远为0

你就非要写成 （double）i 吗

sum改成double类型

结果还是不对哎。。。但是我不按函数调用写，输出就对了。。。为啥啊
#include <stdio.h>
double fun(int m);
int main()
{
int m;
scanf("%d",&m);
printf("The result is %lf\n",fun(m));
return 0;
}
double fun(int m){
double t,i,sum;
if(m>1){
for(i=2;i<=m;i++){
sum=sum+1/i;
} t=1-sum;}
else t=1;
return t;
}