#include <iomanip>
#include <iostream>
#include <Windows.h>
using namespace std;
template <class T>
void sort(int n, T a[]){
for(i=0;i<n-1;i++){
for(j=i;j<n;j++){
if(a[i] > a[j]){
temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
}
void main(){
int size;
cout<<" Please input the amount of numbers you would like to sort"<<endl;
cin>>size;
int Amta[size];
for(int i=0; i<size; i++){
cout<<"Please enter the "<<size+1<< "number";
cin>>Amta[i];
}
Sleep(100000);
}
I am trying to get the how many numbers the user would like to input from the user and store it in the variable size.
But when I initialize array Amta[size] I get the following compile errors:
Expression must have constant value
and
C2057: expected constant expression" compile error.
You can't enter a non-constant value between the brackets when you declare your array:
int Amta[size];
Since you're getting size from the user, the compiler can't tell ahead of time how much memory it needs for Amta. The easiest thing to do here (especially for an exercise) is to just choose a relatively large value and make that the constant allocation, like:
int Amta[1024];
And then if you want to be careful (and you should) you can check if (size > 1024) and print an error if the user wants a size that's beyond the pre-allocated bounds.
If you want to get fancy, you can define Amta with no pre-set size, like int *Amta; and then you allocate it later with malloc:
Amta = (int *)malloc(sizeof(int) * size);
Then you must also free Amta later, when you're done with it:
free(Amta);
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