为什么C++代码中的fun()函数返回值为10？

Oh, this is a super common gotcha with C++ integer literals that trips up a lot of folks! Let me break down exactly what's happening here.

First, let's look at your code again for context:

#include<bits/stdc++.h>
using namespace std;
int fun() {
    int d = 0012;
    return d;
}
int main() {
    cout << fun() << "\n";
    return 0;
}

The Core Issue: Octal Integer Literals #

In C++, any integer literal that starts with a leading 0 (and isn't followed by x/X for hexadecimal) is treated as an octal (base-8) number, not the decimal (base-10) number you might expect.

So when you write 0012:

• The leading zeros are just syntax to signal octal format (they don't add any value)
• We convert the octal digits 12 to decimal:
  • 1 * 8^1 + 2 * 8^0 = 8 + 2 = 10

That's why the variable d ends up holding the value 10, which is what gets returned and printed.

How to Fix It #

If you want to store the decimal value 12, just write it without leading zeros:

int d = 12; // This will store the decimal 12 as you'd expect

Quick Recap of Integer Literal Formats in C++ #

• Decimal: No leading zeros (e.g., 12, 456)
• Octal: Leading 0 (e.g., 012, 056)
• Hexadecimal: Leading 0x or 0X (e.g., 0xC, 0x1A)

内容的提问来源于stack exchange，提问作者Aubdur Rob Anik