Question

我有以下代码：

    //get distances
int a = -1;
int b = -1;
while ((a != 0) && (b != 0)) {
    scanf("%d %d", &a, &b);
    printf("The shortest path from %d to %d is %.1lf meters.\n", a, b, D[a][b]);
}

循环似乎在一次输入后终止，尽管事实上a和b没有被输入为0。

即：

0 2
The shortest path from 0 to 2 is 237.7 meters.

不确定为什么会这样做所以任何帮助都会受到赞赏。

然后终止

需要时的完整代码

#include <stdio.h>
#include <stdlib.h>
#define INF 41e6

double** new2Ddouble(int n, int m);
void free2D(double **a);

int main() {
    //get # of nodes
    int size = 0;
    scanf("%d", &size);

    //create matrix
    double **D = new2Ddouble(size, size);

    //fill with inf for D[i][j]
    int i;
    int j;
    for(i=0; i<size; i++) {
        for(j=0; j<size;j++){
            D[i][j] = INF;
        }
    }

    //fill D[i][i] with INF
    for(i=0;i<size;i++) D[i][i] = INF;

    int exit = 0;
    int I;
    int J;
    double d;
    while(exit != 1) {
        //populate values in matrix
        scanf("%d %d %lf", &I, &J, &d);
        if(I == 0 && J == 0 && d == 0){
            //we can exit
            exit = 1;
        } else {
            D[I][J] = d;
        }
    }

    //calculate distances
    /* Floyd-Warshall Algorithm */
    int k;
    for (k=0; k<size; ++k)
        for (i=0; i<size; ++i)
            for (j=0; j<size; ++j)
                if (D[i][k]+D[k][j] < D[i][j])
                    D[i][j] = D[i][k]+D[k][j];

    exit = 0;
    //get distances
    int a = -1;
    int b = -1;
    while ((a != 0) && (b != 0)) {
        scanf("%d %d", &a, &b);
        printf("The shortest path from %d to %d is %.1lf meters.\n", a, b, D[a][b]);
    }
    return 0;
}


double** new2Ddouble(int n, int m) {
    int i;
    double **ret = (double**) malloc(n*sizeof(double*));
    double *a = (double*) malloc(n*m*sizeof(double));
    for (i=0; i<n; ++i) ret[i] = &a[i*m];
    return ret;
}

void free2D(double **a) { free(a[0]); free(a); }

Answer 1

scanf("%d %d", &a, &b);

在此行中，您分别扫描a和b 0和2的值。因此，一旦将新值扫描到这些变量，那么您的while条件将失败，因为您已显示a为0，并且永远不会检查第二个条件，因为0 && (0|1) = 0

一旦a=0条件失败并退出循环。

Answer 2

while ((a != 0) && (b != 0))

此循环将在a 或 b为零时终止，并在a 和 b时继续不是零。您输入a为0，并且条件a!=0失败，您的循环终止。

Answer 3

there are a couple of problems with this line:

scanf("%d %d", &a, &b);

Note: scanf does not automatically consume white space, 
   so on the second iteration
   the next input char is a newline.
   so scanf fails
Note: all I/O statements 
   should have the returned value checked
   to assure success of the operation

to fix those items, write it like so:

if( 2 != scanf(" %d %d", &a, &b) )
{
    perror( "scanf failed" );
    exit( EXIT_FAILURE );
}

the leading ' ' in the format string causes leading 
white space, like the newline, to be consumed
the 'if' checks the returned value from
scanf to assure that all the input conversions were successful

在C中提前终止的简单while循环问题

3 个答案: