C:在本例中跳过printf和scanf语句。

Question

我想制作2行计算器程序，但是在第一个printf之后，它跳过了第二个语句。

#include <stdio.h>
int main ()
{
    int a,b,c,d,abcd;
    int e,f,g,h,efgh;
    
    printf("1. "); 
    scanf("(%d+%d)x(%d-%d)",&a,&b,&c,&d);
    printf("2. "); 
    scanf("(%d+%d)x(%d-%d)",&e,&f,&g,&h);
    
    abcd=(a+b)*(c-d);
    efgh=(e+f)*(g-h);
    
    printf("%d %d %d",abcd,efgh);
    return 0;
}

我想让这个节目像这样：

input
 1. (1+2)x(3-4)
 2. (5+6)x(7-8)

output
-3 -11

Answer 1

我强烈建议您使用fgets读取一行输入，然后使用sscanf解析所需的信息。您还需要检查sscanf的返回值，以确保读取了所有四个值。在这种情况下，我成功地输入了打破无限循环(for (;;) { ... })所需的输入。

还请注意，所有变量都不需要在函数开始时声明。相反，可以在其使用点声明和初始化它们。在这个场景中，我在稍后的abcd中声明了main和efgh。

#include <stdio.h>

#define LINE_LEN 256

int main ()
{
    int a, b, c, d;
    int e, f, g, h;

    char line[LINE_LEN] = {0};

    for (;;) {
        printf("1. ");
        fgets(line, LINE_LEN, stdin);
        if (sscanf(line, "(%d+%d)x(%d-%d)", &a, &b, &c, &d) == 4)
            break;
 
        fprintf(stderr, "Bad input. Try again.");
    }  

    for (;;) {
        printf("1. ");
        fgets(line, LINE_LEN, stdin);
        if (sscanf(line, "(%d+%d)x(%d-%d)", &e, &f, &g, &h) == 4)
            break;
 
        fprintf(stderr, "Bad input. Try again.");
    } 

    int abcd = (a + b) * (c - d);
    int efgh = (e + f) * (g - h);

    printf("%d\n", abcd);
    printf("%d\n", efgh);

    return 0;
}

Answer 2

通过一些建议，here，我复制了从C中的stdin中读取的函数，并修改了您的示例。这是处理来自C应用程序的stdin的一种安全方法。

此外，我尝试了您的代码，并且我认为错误的是，一些换行符在stdin中被“困住”，因此第二个scanf会在您编写任何东西之前返回。所以这个解决方案。

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

#define OK       0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
    int ch, extra;

    // Get line with buffer overrun protection.
    if (prmpt != NULL) {
        printf ("%s", prmpt);
        fflush (stdout);
    }
    if (fgets (buff, sz, stdin) == NULL)
        return NO_INPUT;

    // If it was too long, there'll be no newline. In that case, we flush
    // to end of line so that excess doesn't affect the next call.
    if (buff[strlen(buff)-1] != '\n') {
        extra = 0;
        while (((ch = getchar()) != '\n') && (ch != EOF))
            extra = 1;
        return (extra == 1) ? TOO_LONG : OK;
    }

    // Otherwise remove newline and give string back to caller.
    buff[strlen(buff)-1] = '\0';
    return OK;
}

int main ()
{
    int a,b,c,d,abcd;
    int e,f,g,h,efgh;
    
    int rc;
    char buff[20];
    
    rc = getLine ("1. ", buff, sizeof(buff));
    if (rc == NO_INPUT) {
        // Extra NL since my system doesn't output that on EOF.
        printf ("\nNo input\n");
        return 1;
    }

    if (rc == TOO_LONG) {
        printf ("Input too long [%s]\n", buff);
        return 1;
    }
    
    sscanf(buff, "(%d+%d)x(%d-%d)",&a,&b,&c,&d);
    
    rc = getLine ("2. ", buff, sizeof(buff));
    if (rc == NO_INPUT) {
        // Extra NL since my system doesn't output that on EOF.
        printf ("\nNo input\n");
        return 1;
    }

    if (rc == TOO_LONG) {
        printf ("Input too long [%s]\n", buff);
        return 1;
    }
    
    sscanf(buff, "(%d+%d)x(%d-%d)",&e,&f,&g,&h);
    
    printf("Check \n");
    printf("%d %d %d %d\n" ,a,b,c,d);
    printf("%d %d %d %d\n" ,e,f,g,h);
    
    abcd=(a+b)*(c-d);
    efgh=(e+f)*(g-h);
    
    printf("Results: %d %d \n",abcd,efgh);
    return 0;
}

Answer 3

扫描是错误的。扫描(“%typeOfFirstVariable%scanf.”，&firstVariable，&secondVariable.)；

int a,b,c,d,abcd;
    int e,f,g,h,efgh;
    
    printf("1. "); 
    scanf("%d %d %d %d)",&a,&b,&c,&d);
    
    ...
    return 0;