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从‘void*’到‘int’的转换损失精度

在CentOS6.2 64位下编译一下代码,不通过,提示 

./11_2.cpp: In function ‘int main(int, char**)’:
./11_2.cpp:28: 错误:从‘void*’到‘int’的转换损失精度
./11_2.cpp:31: 错误:从‘void*’到‘int’的转换损失精度

 1 #include <unistd.h> 2 #include <cstdio> 3 #include <pthread.h> 4  5 using namespace std; 6  7 void *thr_fn1(void *arg) 8 { 9     printf("thread 1 returning\n");10     return (void*)1;11 }12 13 void *thr_fn2(void *arg)14 {15     printf("thread 2 exiting\n");16     pthread_exit((void*)2);17 }18 19 int main(int argc, char **argv)20 {21     pthread_t tid1, tid2;22     void *tret;23 24     pthread_create(&tid1, NULL, thr_fn1, NULL);25     pthread_create(&tid2, NULL, thr_fn2, NULL);26 27     pthread_join(tid1, &tret);28     printf("thread 1 exit code %ld\n", (long)tret);29 30     pthread_join(tid2, &tret);31     printf("thread 2 exit code %ld\n", (long)tret);32 33     return 0;34 }

既然提示精度损失,那么看一下各自的精度即可:

 1 #include <iostream> 2  3 using namespace std; 4  5 int main(int argc, char **argv) 6 { 7     cout << sizeof(int) << endl; 8     cout << sizeof(long) << endl; 9     cout << sizeof(void*) << endl;10 11     return 0;12 }

执行结果:

4
8
8

好吧,确实是精度损失了,从4个字节转换为8个字节。但是问题来了,为什么在64位下,指针是8个字节呢?

从‘void*’到‘int’的转换损失精度