[英]No output from cout when calling void function
I'm new to C++ so bear with me. 我是C ++的新手,请多多包涵。
I am trying to create a histogram from certain parameters (interval size, length of array containing quantities of numbers, highest number yadayada). 我正在尝试根据某些参数(间隔大小,包含数字数量的数组的长度,最高数量的yadayada)创建直方图。
The details are irrelevant and a problem for myself to fiddle with, although I think I got the correct formula in my function. 尽管我认为我的函数中使用了正确的公式,但这些细节无关紧要,也是我自己不易解决的问题。
When I assign variables from the C++ IO "cin" I can output those with the "cout" call, however, when I call my histogram function, also containing "cout" instructions, nothing gets printed. 当我从C ++ IO的“ cin”分配变量时,可以通过“ cout”调用输出变量,但是,当我调用还包含“ cout”指令的直方图函数时,则不会打印任何内容。
My code: 我的代码:
#include <iostream>
#include <cmath>
using namespace std;
void histogram(int l, int n, int k, int *a)
{
int quantity = 0;
for (int i = 1; i <= l; i++)
{
for (int j = 0; i < n; j++)
{
if (a[j] >= (i-1) * k || a[j] <= i * k)
{
quantity++;
}
}
cout << (i-1) * k + ": " + quantity << endl;
quantity = 0;
}
}
int main()
{
int l,n,k;
int *a;
a = new int[n];
cin >> l >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
k = ceil((double)a[0]/l);
// cout << k;
histogram(l,n,k,a);
return 0;
}
There might be something funky going on with this line and string concatenation: cout << (i-1) * k + ": " + quantity << endl;
这行和字符串的连接可能有些时髦:
cout << (i-1) * k + ": " + quantity << endl;
You might try rewriting as cout << ((i-1) * k) << ": " << quantity << endl;
您可以尝试将其重写为
cout << ((i-1) * k) << ": " << quantity << endl;
just to ensure that things are adding and concatenating correctly. 只是为了确保事物正确地添加和连接。
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