C 程序在运行时冻结

Question

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

int main()
{
    int x1, x2, x3, 
        y1, y2, y3, 
        ystr1, ystr2, ystr3,  
        xstr1, xstr2, xstr3, 
        sstr1, sstr2, sstr3;

    printf("Insert.");
    scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);
    ystr1 = y2 - y1;
    xstr1 = x2 - x1;
    sstr1 = sqrt(pow(xstr1, 2) + pow(ystr1, 2));
    ystr2 = y3 - y2;
    xstr2 = x3 - x2;
    sstr2 = sqrt(pow(xstr2, 2) + pow(ystr2, 2));
    ystr3 = y3 - y1;
    xstr3 = x3 - x1;
    sstr3 = sqrt(pow(xstr1, 2) + pow(ystr1, 2));
    printf("Print %d, %d, %d", sstr1, sstr2, sstr3);
    return 0;
}

For some reason this code freezes every time I enter in a single digit. I really don't know what might be causing this. Is it the too many ints?

Answer 1

The scanf function expects pointers as arguments so that it can modify the variables you're reading into. Remember that all arguments in C are passed by value. If we want to modify something with a function, we need to pass its address in memory (a pointer).

This is made apparent by the warnings your code generates even without explicit warnings enabled.

test.c: In function ‘main’:
test.c:14:13: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int’ [-Wformat=]
     scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);
            ~^
test.c:14:15: warning: format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘int’ [-Wformat=]
     scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);
              ~^
test.c:14:17: warning: format ‘%d’ expects argument of type ‘int *’, but argument 4 has type ‘int’ [-Wformat=]
     scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);
                ~^
test.c:14:19: warning: format ‘%d’ expects argument of type ‘int *’, but argument 5 has type ‘int’ [-Wformat=]
     scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);
                  ~^
test.c:14:21: warning: format ‘%d’ expects argument of type ‘int *’, but argument 6 has type ‘int’ [-Wformat=]
     scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);
                    ~^
test.c:14:23: warning: format ‘%d’ expects argument of type ‘int *’, but argument 7 has type ‘int’ [-Wformat=]
     scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);

So instead of:

scanf("%d%d%d%d%d%d", x1, y1, x2, y2, x3, y3);

You'd want to write:

scanf("%d%d%d%d%d%d", &x1, &y1, &x2, &y2, &x3, &y3);

The & operator returns the address of a variable.

It's also important to note that scanf returns an int value. This value indicates the number of values read. You should probably check that scanf("%d%d%d%d%d%d", &x1, &y1, &x2, &y2, &x3, &y3); returns 6 .

if (scanf("%d%d%d%d%d%d", &x1, &y1, &x2, &y2, &x3, &y3) == 6) {
   ...
}

Another thing you may wish to do is set initial values for your variables. If scanf fails, you'd still have that initial value.