Question

#import <Foundation/Foundation.h>

int main (int argc, const char * argv[])
{
    NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];

    NSString *sg, *ge, *pr, *ln;
    NSString *done = @"";
    int actn = 0; 
    NSError *error;

    printf("Welcome to Your Playlist\n");

    while(actn != 6)
    {
        printf("\nWhat would you like to do?");
        printf("\n1. Enter a New Song, Performer, Genre, and the Length of the Song");
        printf("\n2. Show all Songs in Your Playlist");
        printf("\n3. Show all Songs in Your Playlist and Sort by Song Title");
        printf("\n4. Show all Songs in Your Playlist and Sort by Genre");
        printf("\n5. Show all Songs in Your Playlist and Sort by Performer");
        printf("\n6. Save and Quit\n");
        scanf("%d", &actn);

        if(actn == 1)
        {
            while([done isEqualToString:@"No"] == false)
            {
                printf("What is the Song Title? ");
                scanf("%@", &sg);

                printf("Who is the Performer? ");
                scanf("%@", &pr);

                printf("What is the Genre of this Song? ");
                scanf("%@", &ge);

                printf("What is the length of the Song? ");
                scanf("%@", &ln);

                printf("Would you like to quit(Yes or No)? ");        
                scanf("%@", &done);
        }   
    }  

    if(actn == 2)
    {
        NSString *contents = [NSString stringWithContentsOfFile:@"E:/Playlist" encoding:NSUTF8StringEncoding error:&error];

        if(error) 
        { 
            NSLog(@"ERROR while loading from file: %@", error);
        }

        [contents writeToFile:@"E:/Playlist" atomically:YES encoding:NSUnicodeStringEncoding error:&error];
        printf("%@",contents);
    }


    [pool release];
    return 0;
}

它允许输入一个数字，我从键入1开始。它要求我输入一首歌的标题。之后，它将无限制地在while循环中打印下几行而不要求我输入。我该如何解决这个问题？

Answer 1

您使用的是错误的格式说明符：

scanf("%@", &sg);
       ^^

由于scanf()是C运行时库函数，因此不能使用Foundation类型。相反，您需要使用char数组和%s说明符，但OSX可能支持%ms说明符，因为它通过{{1}分配char *更安全所以不能轻易溢出。

目标C：如何修复scanf（）跳过线

1 个答案: