两个线程打印到100，帮我看下为啥会打印到101

#include
#include
#include

int N = 100;
int n = 0;
pthread_mutex_t mutex;

void *func1()
{
while (n < N+1)
{
pthread_mutex_lock(&mutex);
printf ("%d 1\n", n);
n++;
pthread_mutex_unlock(&mutex);
}
if (n == N+1)
pthread_exit(NULL);
}

void *func2()
{
while (n < N+1)
{
pthread_mutex_lock(&mutex);
printf ("%d 2\n", n);
n++;
pthread_mutex_unlock(&mutex);
}

if (n == N+1)
    pthread_exit(NULL);

}

int main(void)
{
pthread_t pid1, pid2;
pthread_mutex_init(&mutex, NULL);

if ((pthread_create(&pid1, NULL, func1, NULL)) != 0)
    printf ("pthread create fail\n");

if ((pthread_create(&pid2, NULL, func2, NULL)) != 0)
        printf ("pthead create fail\n");

pthread_join(pid1, NULL);
pthread_join(pid2, NULL);

return 0;

}

输出如下：
0 2
1 2
2 2
3 2
4 2
5 2
6 2
7 2
8 2
9 2
10 2
11 2
12 2
13 2
14 2
15 2
16 2
17 2
18 2
19 2
20 2
21 2
22 2
23 2
24 2
25 2
26 2
27 2
28 2
29 2
30 2
31 2
32 2
33 2
34 2
35 2
36 2
37 2
38 2
39 2
40 2
41 2
42 2
43 2
44 2
45 2
46 2
47 2
48 2
49 2
50 2
51 2
52 2
53 2
54 2
55 2
56 2
57 2
58 2
59 2
60 2
61 2
62 2
63 2
64 2
65 2
66 2
67 2
68 2
69 2
70 2
71 2
72 2
73 2
74 2
75 2
76 2
77 2
78 2
79 2
80 2
81 2
82 2
83 2
84 2
85 2
86 2
87 2
88 2
89 2
90 2
91 2
92 2
93 2
94 2
95 2
96 2
97 2
98 2
99 2
100 2
101 1

还有一个问题，当我把pthread_mutex_init(&mutex, NULL);放到main函数外面时，编译一种报错
报错代码如下：
a.c:8:20: error: expected declaration specifiers or ‘...’ before ‘&’ token
pthread_mutex_init(&mutex, NULL);
^
In file included from /usr/include/time.h:37:0,
from /usr/include/pthread.h:24,
from a.c:3:
a.c:8:28: error: expected declaration specifiers or ‘...’ before ‘(’ token
pthread_mutex_init(&mutex, NULL);

while (n < N+1) 让它小于N

当线程1 n = 100时，完成while逻辑判断，这时还没有进入 mutex锁，而线程2进入了mutex锁，并执行了n++，这时n=101，而线程1已经在n=100时完成了while逻辑判断，从而print出了101
所以mutex锁应该把while的逻辑判断也包含进去

我的看法，仅供参考。C的对函数的判断是通过函数名+参数来区别的，列在main之外，编译器在认为你想重写pthread_mutex_init函数，但是你的参数没写对，所以会报错。