我正在尝试学习 C 语言的读/写程序,我在网上找到了一个例子,我认为我应该开始学习。当我尝试编译代码时,我收到一些警告和一个错误,如下所示:
read.c:21:8: error: conflicting types for ‘write’
void * write(void *temp) {
^
In file included from read.c:11:0:
/usr/include/unistd.h:369:16: note: previous declaration of ‘write’ was here
extern ssize_t write (int __fd, const void *__buf, size_t __n) __wur;
^
read.c: In function ‘write’:
read.c:25:4: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
ret=pthread_rwlock_wrlock(&rwlock);
^
read.c: In function ‘write_2’:
read.c:45:4: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
ret=pthread_rwlock_wrlock(&rwlock);
^
read.c: At top level:
read.c:106:1: warning: return type defaults to ‘int’ [-Wimplicit-int]
main() {
我对 C 编程还很陌生,我不知道如何解决这些警告和错误,因为我只是在不太了解该语言的情况下跳入这个主题。这是我正在尝试编译的代码。
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <unistd.h>
/*
From the output we can see that the two writes were executed one after the other. But in case of reads, even though read_1 had not unlocked the rwlock, read_2 was allowed into the critical section and read the file. That shows us that multiple readers are allowed but only one writer is allowed into the critical section.
*/
pthread_rwlock_t rwlock; // allows multiple readers to access the resource, but only one reader at any given time.
void * write(void *temp) {
char *ret;
FILE *file1;
char *str;
ret=pthread_rwlock_wrlock(&rwlock);
printf("\nFile locked, please enter the message \n");
str=(char *)malloc(10*sizeof(char));
file1=fopen("temp","w");
scanf("%s",str);
fprintf(file1,"%s",str);
fclose(file1);
pthread_rwlock_unlock(&rwlock);
printf("\nUnlocked the file you can read it now \n");
return ret;
}
void * write_2(void *temp) {
char *ret;
FILE *file1;
char *str;
sleep(3);
ret=pthread_rwlock_wrlock(&rwlock);
printf("\nFile locked, please enter the message \n");
str=(char *)malloc(10*sizeof(char));
file1=fopen("temp","a");
scanf("%s",str);
fprintf(file1,"%s",str);
fclose(file1);
pthread_rwlock_unlock(&rwlock);
printf("\nUnlocked the file you can read it now \n");
return ret;
}
void * read_1(void *temp) {
char *ret;
FILE *file1;
char *str;
sleep(5);
pthread_rwlock_rdlock(&rwlock);
printf("\n1 Opening file for reading\n");
file1=fopen("temp","r");
str=(char *)malloc(10*sizeof(char));
fscanf(file1,"%s",str);
printf("\nMessage from file is %s \n",str);
sleep(3);
fclose(file1);
printf("\nUnlocking rwlock\n");
pthread_rwlock_unlock(&rwlock);
return ret;
}
void * read_2(void *temp) {
char *ret;
FILE *file1;
char *str;
sleep(6);
pthread_rwlock_rdlock(&rwlock);
printf("\n2 Opening file for reading\n");
file1=fopen("temp","r");
str=(char *)malloc(10*sizeof(char));
fscanf(file1,"%s",str);
printf("\nMessage from file is %s \n",str);
fclose(file1);
pthread_rwlock_rdlock(&rwlock);
return ret;
}
main() {
pthread_t thread_id,thread_id1,thread_id3,thread_id4;
pthread_attr_t attr;
int ret;
void *res;
pthread_rwlock_init(&rwlock,NULL);
ret=pthread_create(&thread_id,NULL,&write,NULL);
ret=pthread_create(&thread_id1,NULL,&read_1,NULL);
ret=pthread_create(&thread_id3,NULL,&read_2,NULL);
ret=pthread_create(&thread_id4,NULL,&write_2,NULL);
printf("\n Created thread");
pthread_join(thread_id,&res);
pthread_join(thread_id1,&res);
pthread_join(thread_id3,&res);
pthread_join(thread_id4,&res);
pthread_rwlock_destroy(&rwlock);
}
问题是为什么这段代码会出现这些警告和错误?
附注不知道为什么,但是当我在收到这些警告和错误后尝试运行它时,它仍然按照我想象的方式运行。这很奇怪。感谢您的阅读。
最佳答案
write
是系统调用的名称。该函数的声明位于 <unistd.h>
中您已将其包含到您的 C 程序中。
您的 C 程序继续工作,因为您没有使用实际 write
系统调用; C 标准库使用它,但它链接到其他 write
静态运行。
至于其他警告,
main()
根本不是正确的C。它需要具有 int main(void)
的原型(prototype)或int main(int argc, char *argv[])
.
对于pthread_*
您必须的功能#include <pthread.h>
.
返回值pthread_rwlock_wrlock
是不是指针,而是 int
,因此您必须将其分配给 int
类型的对象,但您将其分配给 ret
其类型为 char *
总而言之,您应该在 C 编译器中打开所有通常有用的诊断 ( -Wall
),并将每个警告视为错误 ( -Werror
)。
最后,所有编译和链接命令行都应该有 -pthread
在他们身上标记。
关于冲突类型错误和整数指针,没有强制转换警告 C,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53717915/