Why Doesn't `vector.size() - 1` Return -1 in C ?

Vector Size Anomalies in C

In C , the vector container's size() method returns the number of elements in the vector. However, perplexing behavior arises when the vector is empty, as demonstrated in the following code snippet:

#include <vector>
#include <iostream>

using namespace std;

int main() {
    vector<int> value;

    cout <p><strong>Why is the Second Output Not -1?</strong></p>
<p>The first cout statement correctly prints 0, indicating that the vector is empty. However, the second statement surprisingly outputs a very large positive number, not the expected -1. This occurs because vector::size() returns an unsigned integer of type size_t. Unsigned integers, as the name suggests, can't represent negative numbers.</p>
<p>Therefore, when a size_t value is subtracted from itself, it essentially wraps around to the maximum representable number for that type. In this case, size_t is typically a 64-bit integer, so the maximum value it can hold is 18446744073709551615. This value is what's printed as the output of the second cout statement.</p>
<p><strong>What Happens During the Second cout?</strong></p>
<p>The following sequence of calculations explains the behavior behind the second cout:</p>
<ol>
<li>value.size() returns 0, an unsigned integer of type size_t.</li>
<li>value.size() - 1 computes 0 - 1, resulting in 18446744073709551615 (the maximum representable number for size_t).</li>
<li>cout prints this maximum value as 18446744073709551615.</li>
</ol></int></iostream></vector>

The above is the detailed content of Why Doesn't `vector.size() - 1` Return -1 in C ?. For more information, please follow other related articles on the PHP Chinese website!