I have two “equivalent” programs in C and in C++ that have different outputs.
They should print step by step all the powers of a “base” between 0 and “exp”
test.c is:
#include <stdio.h>
int power(int b, int e) {
int p = 1;
for (int i = 0; i < e; i++) {
printf("%d^%d = %dn", b, i, p);
p = p * b;
}
return p;
}
int main() {
int base = 2, exp = 3;
printf("%d^%d = %dn", base, exp, power(base, exp));
return 0;
}
output:
2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
test.cpp is:
#include <iostream>
using namespace std;
int power(int b, int e) {
int p = 1;
for (int i = 0; i < e; i++) {
cout << b << "^" << i << " = " << p << endl;
p = p * b;
}
return p;
}
int main() {
int base = 2, exp = 3;
cout << base << "^" << exp << " = " << power(base, exp) << endl;
return 0;
}
output:
2^3 = 2^0 = 1
2^1 = 2
2^2 = 4
8
Can anybody please tell me why does this happen?
Why does the “cout” print the first part before executing the function?
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2
cout is evaluated left-to-right. It simply prints everything it sees, when it sees it. So, in your case, your function is called only after printing the first part of the line.
printf, on the other hand, needs to evaluate everything to fill itself in. So, it runs your function first and then prints.
If you want to get this behavior in C++, it probably makes the most sense to define your output as a string, fill in your output, and then print that all at once, such as:
#include <iostream>
#include <sstream>
using namespace std;
int power(int b, int e) {
int p = 1;
for (int i = 0; i < e; i++) {
cout << b << "^" << i << " = " << p << endl;
p = p * b;
}
return p;
}
int main() {
stringstream out;
int base = 2, exp = 3;
out << base << "^" << exp << " = " << power(base, exp);
cout << out.str() << endl;
return 0;
}
This code has the exact same output as your C code.
This is how operator << works. First one is evaluated it returns reference to std::cout then next operator << is evaluated and so one. So this C++ code is equivalent to:
int main()
{
int base = 2, exp = 3;
cout << base;
cout << "^";
cout << exp;
cout << " = ";
cout << power(base, exp);
cout << endl;
return 0;
}
So main prints something before calling power and prints something after calling power. Here is some tool which expands C/C++ expressions which shows this to some extent.
In C code all arguments must be evaluated first before main can print, so power must finish printing before printing in main can be done.
Note C++23 has introduced std::print which is type safe as std::cout and behaves like printf in this scenario:
#include <print>
int power(int b, int e)
{
int p = 1;
for (int i = 0; i < e; i++) {
std::print("{}^{} = {}n", b, i, p);
p = p * b;
}
return p;
}
int main()
{
int base = 2, exp = 3;
std::print("{}^{} = {}n", base, exp, power(base, exp));
return 0;
}
https://godbolt.org/z/vEeMP5nbP