c - 我是 C 新手，正在测试代码，并且在使用函数时遇到困难

Question

这是代码，但我收到错误，实时参数太少，并且字符之前的预期表达式。

我正在尝试编写一个方法，然后通过该方法传递信息(输入)并获取值(输出)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>



char realtime(char input[20], char output[25]){

    //char input[20];
    //char output [25];
    char year[5];
    char month[3];
    char day[3];
    char hour[3];
    char min[3];
    char sec[3];
    strncpy(year, input, 4);
    year[4] = '\0';
    strncpy(month, input + 4, 2);
    month[2] = '\0';
    strncpy(day, input + 6, 2);
    day[2] = '\0';
    strncpy(hour, input + 8, 2);
    hour[2] = '\0';
    strncpy(min, input + 10, 2);
    min[2] = '\0';
    strncpy(sec, input + 12, 2);
    sec[2] = '\0';
    sprintf(output, "%s-%s-%sT%s:%s:%s", year, month, day, hour, min, sec);
    return 0;
}

int main(){
    char input = "20181204193456";
    realtime( char input[20], char output[25]);
    printf("Parsed Date %s", output);
}

Answer 1

您需要在main中定义一个变量来保存输出，并且对realtime的定义和调用不正确。也许这样会更好:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char realtime(char *input, char *output){

    //char input[20];
    //char output [25];
    char year[5];
    char month[3];
    char day[3];
    char hour[3];
    char min[3];
    char sec[3];
    strncpy(year, input, 4);
    year[4] = '\0';
    strncpy(month, input + 4, 2);
    month[2] = '\0';
    strncpy(day, input + 6, 2);
    day[2] = '\0';
    strncpy(hour, input + 8, 2);
    hour[2] = '\0';
    strncpy(min, input + 10, 2);
    min[2] = '\0';
    strncpy(sec, input + 12, 2);
    sec[2] = '\0';
    sprintf(output, "%s-%s-%sT%s:%s:%s", year, month, day, hour, min, sec);
    return 0;
}

int main(){
    char input[] = "20181204193456";
    char output[25];

    realtime(input, output);
    printf("Parsed Date %s", output);

    return 0;
}