c - 错误 : called object type 'ssize_t' (aka 'long' ) is not a function or function pointer

Question

我在 C 编程方面不是很有经验，我正在尝试使用管道编写程序。问题是，当我尝试编译它时，它给了我一个我无法理解的奇怪错误，即使通过阅读 Stackoverflow 中的其他类似问题也是如此。

错误是下一个:

ejercicio.c:91:8: error: called object type 'ssize_t' (aka 'long') is not a function or function pointer
                    read(tuberiaCom[0], login, sizeof(int));
                    ~~~~^
ejercicio.c:103:8: error: called object type 'ssize_t' (aka 'long') is not a function or function pointer
                        read(tuberiaCom[0], *checkuser, 256);

受影响的代码是这个:

int main(){
    int tuberiaCom[2];
    int fduser;
    int login = 0;
    char username[256];
    char answer[1024];
    char checkuser[256];
    char command[64];
    char program[16];
    char options[48];
    char *line = NULL;
    pid_t familia;
    size_t len = 0;
    ssize_t read;

    if(pipe(tuberiaCom) == -1){
        fprintf(stderr, "Pipe error\n");
        exit(1);
    }

    switch(familia = fork()){
        case -1:
            fprintf(stderr, "Fork error\n");
            exit(1);
            break;
        case 0:
            scanf("Please, write your username %s", username);

            write(tuberiaCom[1], &username, strlen(username));
            read(tuberiaCom[0], login, sizeof(int));    /////////Line 91

            if(login == 1){
                printf("User %s exists, you can continue", username);
            }
            else{
                printf("There is no user called %s, the program will finnish", username);
                exit(1);
            }

            break;
        default:
            read(tuberiaCom[0], *checkuser, 256);    /////////Line 103
            fduser = open("users.txt", O_RDONLY);

            while ((read = getline(&line, &len, fduser)) != -1){    
                if(*line == *checkuser){
                    login = 1;
                    break;
                }
            }

            write(tuberiaCom[1], &login, sizeof(int));

            break;
    }
    ......

Answer 1

您有一个名为 read 的变量，当您调用 read(2) 时，它会使编译器与函数 read(2) 混淆。将您的变量 read 重命名为其他名称。