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akko
Asked: 2020-01-14 00:47:00 +0800 CST 2020-01-14 00:47:00 +0800 CST

Help with usleep function in Linux C++

  • 772

I have a function that pauses the console for a few seconds and then continues to the next function:

void esperar(double segundos){
    if(segundos < 0) return;
    #ifdef _WIN32
        segundos *= 1000;
        int goal = (int) segundos + clock();
        while(goal > clock());
    #else
        segundos *= 1000000;
        int goal = (int) segundos;
        usleep(goal);
    #endif
}

Assuming I want to print the numbers 1 to 10 with a for loop I get two different answers depending on whether there is a line break or not.

for(int i = 0; i < 10; ++i){
    cout << i + 1 << " ";
    esperar(0.1);
}

In the first loop I leave a space after each number, but this loop waits a tenth of a second (0.1 * 1000000) for each cout in the loop, for a total of 1 second. Only until the 10 cycles of 0.1 seconds are finished is the string printed.

1 2 3 4 5 6 7 8 9 10

In the second loop I leave a line break between each number.

for(int i = 0; i < 10; ++i){
    cout << i + 1 << endl;
    esperar(0.1);
}

But this has a different (and desired) behavior. After printing the first cout the console waits 0.1 seconds to print the second cout, waits 0.1 seconds and prints the next one....

1
//0.1 segundos
2
//0.1 segundos
3
//0.1 segundos
4
//0.1 segundos
5
//0.1 segundos
[...]

I have the following libraries included

#include<time.h>
#include<unistd.h>

Is there a way to correct this?

c++

1 Answers

  1. Best Answer

    In the first version you use a space as a separator:

    for(int i = 0; i < 10; ++i){
    cout << i + 1 << " ";
    esperar(0.1);
}

    What is happening here is that you are accumulating in the output buffer a sequence of numbers. Accumulating them in the output buffer does not imply that they are being displayed on the screen. The buffer is not synchronous with the console.

    At a given moment a synchronization will be forced (because the program ends, a call to a data input method is made, ...) and then the content of the buffer will be dumped to the screen.

    In this case I assume that the results will be displayed on the screen after a wait of 1 second (0.1 seconds of waiting for 10 numbers to be displayed).

    If we now see the second case:

    for(int i = 0; i < 10; ++i){
    cout << i + 1 << endl;
    esperar(0.1);
}

    You are now calling endl. Although it endlintroduces, as we all know, a line break, this element has another added functionality and that is that it forces the output buffer to be synchronized. This forces that, after entering a number in the buffer, it is dumped to the console.

    For the reason mentioned above, in the second case you observe that between the appearance of each number there is a wait of 0.1 seconds.

    I imagine that what you want is to correct the first case. To force output buffer synchronization you can use std::flush:

    for(int i = 0; i < 10; ++i){
    cout << i + 1 << " " << std::flush;
    esperar(0.1);
}
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