Question

我所提的问题与某些已回答的问题相似，但那里的答案却很特别。这是我的代码

Microsoft.Extensions.Logging.ILogger

我想检查输入是否为负，然后才将其添加到一维数组中。问题是我该怎么做？有任何想法吗？

感谢广告

Answer 1

常规提示

int main()
{   

int i,j,k,sorted; // do not declare loop control variables outside of their respective loops
int A[4][4];
int C[16];
int positive = 0; // for readability's sake please refrain from using long ass variable names for simple control var's

for(i=0;i<4;i++) // idiomatic way is for(int i = 0; i < 4; i++)
{
    for(j = 0; j<4; j++) // for(int j = 0; j<4; j++)
    {
        printf("A[%d][%d]: ", i,j);
        scanf("%d", &A[i][j]);

        if(A[i][j]>0)
        {
            C[positive] = A[i][j];
            printf("C = %d\n"); // missing argument & redundant since you intend to print them afterwards anyway? 
            positive++;
        }       
    }
}
for(j=0;j<4;j++) // this loop is supposed to print all positive numbers?
{
    printf("%d  ", A[positive]); // you are looping through A[0-3][0], which are not necessarily the positive numbers of A.
}
printf("\n");
//printf("Your positive numbers are: ", positive);

现在您的问题就在眼前。代码尝试执行的操作以及您向我们提出的要求是两件事。现在您正在向C数组中添加正值，但是您的问题询问的是负值吗？

按照我认为的方式清理代码。

int main()
{
int A[4][4];
int C[16];
int k = 0;
for (int i = 0; i < 4; i++)
{
    for (int j = 0; j < 4; j++)
    {
        printf("A[%d][%d]: ", i, j);
        scanf("%d", &A[i][j]);
        if (A[i][j] > 0) 
        {
            C[k] = A[i][j];
            k++;
        }
    }
}
printf("positive numbers in matrix A are: ");
for (int i = 0; i < k; i++) // loop from 0 to actual number of positive integers; array size remains at 16 in this case.
{
    printf("%d  ", C[i]);
}
printf("\n");
//printf("Your positive numbers are: ", positive);
printf("\n");
system("pause");
return 0;
}

Answer 2

int main()
{   int i,j,k,sorted;
int A[4][4];
int C[16];
int positive = 0;

for(i=0;i<4;i++)
{
    for(j = 0; j<4; j++)
    {
        printf("A[%d][%d]: ", i,j);
        scanf("%d", &A[i][j]);

        if(A[i][j]>0){
            C[positive] = A[i][j];
            printf("C = %d\n");         
            positive++;
        }       
    }
}
for(j=0;j<positive;j++)
{
    printf("%d  ", C[j]);
}
printf("\n");
//printf("Your positive numbers are: ", positive);

printf("\n");
system("pause");
return 0;

}

在错误的循环结果中，您打印了一个2D数组，但您仅使用1D数组，并且在循环中直到4为止，直到正数，然后在循环中打印了A，但是结果是C否A

仅读取和打印正数组索引

2 个答案: