Why Does `time()` Sometimes Return the Same Value Before and After a Short Task?

Easily Measure Elapsed Time

Question:

When measuring time elapsed in a program using time(), why might the before and after values remain the same?

Explanation:

time() returns the current time in seconds since the Epoch. If the time span to be measured is short, the difference between the before and after values may be too small to be detected.

Solution:

To accurately measure short time spans, consider using alternative methods:

C Structured Time Measurement:

struct timeval startTV, endTV;
gettimeofday(&startTV, NULL); 
// Execute time-consuming tasks
gettimeofday(&endTV, NULL); 

timersub(&endTV, &startTV, &diff);
printf("**time taken = %ld %ld\n", diff.tv_sec, diff.tv_usec);

• timeval structure represents a time value with precision up to microseconds.
• gettimeofday() obtains the current time of day.
• timersub() calculates the difference between two timeval structures.
• The result is reported in seconds (diff.tv_sec) and microseconds (diff.tv_usec).

C 11 Style Precision Time Measurement:

#include <chrono>

std::chrono::steady_clock::time_point begin = std::chrono::steady_clock::now();
// Execute time-consuming tasks
std::chrono::steady_clock::time_point end = std::chrono::steady_clock::now();

std::cout << "Time difference = " << std::chrono::duration_cast<std::chrono::microseconds>(end - begin).count() << "[µs]" << std::endl;
std::cout << "Time difference = " << std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin).count() << "[ns]" << std::endl;

• std::chrono library provides high-precision time measurement.
• std::chrono::steady_clock measures time since an unknown arbitrary point in time.
• The time difference is calculated as a std::chrono::duration object, which can be converted to microseconds or nanoseconds.

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