What costs the extra execution time of the routine in a pthread program?

Question

I wrote four different programs to count total words in two files. These four versions look mostly the same. First three versions use two threads to count and just the orders of three statements are different. The last version uses one thread to count. I will list the different part of each version and the common part first, then the output of each version and my question. different part: ``` // version 1 count_words(&file1); pthread_create(&new_thread, NULL, count_words, &file2); pthread_join(new_thread, NULL); // version 2 pthread_create(&new_thread, NULL, count_words, &file2); count_words(&file1); pthread_join(new_thread, NULL); // version 3 pthread_create(&new_thread, NULL, count_words, &file2); pthread_join(new_thread, NULL); count_words(&file1); // version 4 count_words(&file1); count_words(&file2); ``` common part: (Insert the different part into this common part to make a complete version) ``` #include #include #include #include #incl…

Accepted Answer

It's probably because creation of any thread forces libc to use synchronization in `getc`. This makes this function significantly slower. Following example is for me as slow as version 3:

```
void *skip(void *p){ return NULL; };
pthread_create(&new_thread, NULL, skip, NULL);
count_words(&file1); 
count_words(&file2); 
```

To fix this problem you can use a buffer:

```
for (i = 0; i < N; i++) {
    char buffer[BUFSIZ];
    int read;
    do {
        read = fread(buffer, 1, BUFSIZ, fp);
        int j;
        for(j = 0; j < read; j++) {
            if (!isalnum(buffer[j]) && isalnum(prevc))
                file->words++;
            prevc = buffer[j];
        }
    } while(read == BUFSIZ);
    fseek(fp, 0, SEEK_SET);
}
```

In this solution, IO functions are called rarely enough to make synchronization overhead insignificant. This not only solves problem of weird timings, but also makes it several times faster. For me it's reduction from `0.54s` (without threads) or `0.85s` (with threads) to `0.15s` (in both cases).

What costs the extra execution time of the routine in a pthread program?

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