Why Does Comparing -1 to an Empty Vector's Size Result in 'Greater' in C ?

Why Comparing -1 to a Vector's Size of 0 Results in "Greater"

The Puzzle

In C , the following code snippet produces the unexpected output of "Greater":

#include <iostream>
#include <vector>

int main() {
    std::vector<int> a;
    std::cout << "vector size " << a.size() << std::endl;

    int b = -1;
    if (b < a.size())
        std::cout << "Less";
    else
        std::cout << "Greater";
}

Output:

vector size 0
Greater

Unraveling the Mystery

The reason behind this counterintuitive behavior lies in the type difference between the values being compared. a.size() returns an unsigned integer, representing the non-negative size of the vector. On the other hand, b is a signed integer, holding the negative value -1.

When comparing these two values, C performs an implicit promotion to unsigned. As a result, b is promoted to a large unsigned integer, which is then compared to the unsigned a.size(). The large unsigned value surpasses 0, leading to the "Greater" output.

A Clarifying Example

This behavior can be further illustrated by the following code:

#include <iostream>

int main()
{
    unsigned int a = 0;
    int b = -1;
    std::cout << std::boolalpha << (b < a) << "\n"; 
}

Output:

false

Here, we directly compare an unsigned type (a) with a signed type (b). The negative value b is promoted to a large unsigned integer, which is rightfully considered greater than the unsigned a with a value of 0.

Conclusion

Understanding the type differences and the implicit promotion rules is crucial for working with signed and unsigned values in C . When comparing integers of different types, it's important to consider the way in which they are promoted and whether the resulting comparison is meaningful.

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