如何通过引用pthread启动例程传递顺序计数器？

Question

Below is my C code to print an increasing global counter, one increment per thread.

#include <stdio.h>
#include <pthread.h>

static pthread_mutex_t pt_lock = PTHREAD_MUTEX_INITIALIZER;
int count = 0;

int *printnum(int *num) {
    pthread_mutex_lock(&pt_lock);
    printf("thread:%d ", *num);
    pthread_mutex_unlock(&pt_lock);
    return NULL;
}

int main() {
    int i, *ret;
    pthread_t pta[10];
    for(i = 0; i < 10; i++) {
        pthread_mutex_lock(&pt_lock);
        count++;
        pthread_mutex_unlock(&pt_lock);
        pthread_create(&pta[i], NULL, (void *(*)(void *))printnum, &count);
    }
    for(i = 0; i < 10; i++) {
        pthread_join(pta[i], (void **)&ret);
    }
}

I want each thread to print one increment of the global counter but they miss increments and sometimes access same values of global counter from two threads. How can I make threads access the global counter sequentially?

Sample Output :

thread:2 thread:3 thread:5 thread:6 thread:7 thread:7 thread:8 thread:9 thread:10 thread:10

Edit

Blue Moon's answer solves this question. Alternative approach is available in MartinJames'es comment.

Answer 1

A simple-but-useless approach is to ensure thread1 prints 1 , thread2 prints 2 and so on is to put join the thread immmediately:

pthread_create(&pta[i], NULL, printnum, &count);
pthread_join(pta[i], (void **)&ret);

But this totally defeats the purpose of multi-threading because only one can make any progress at a time.

Note that I removed the superfluous casts and also the thread function takes a void * argument.

A saner approach would be to pass the loop counter i by value so that each thread would print different value and you would see threading in action ie the numbers 1-10 could be printed in any order and also each thread would print a unique value.