Question

您好，这里的初学者需要您的帮助。我的C程序很好，并且只执行了不应使用任何种类的if语句的操作。我认为这样写起来会更容易，因此可以替换if语句。我一直在尝试替换if语句，但是现在卡住了。我可以使用什么代替if语句来产生相同的输出。

该程序应生成一个由30个随机整数组成的序列（介于0和9之间），然后向前和向后打印该序列。然后打印出一个数字，显示序列中0到9之间的每个数字出现了多少次。

这是输出

Here is a sequence of 30 random numbers between 0 and 9:
 3 6 7 5 3 5 6 2 9 1 2 7 0 9 3 6 0 6 2 6 1 8 7 9 2 0 2 3 7 5

Printing them backwards, that's:
  5 7 3 2 0 2 9 7 8 1 6 2 6 0 6 3 9 0 7 2 1 9 2 6 5 3 5 7 6 3

There were 3 0's
There were 2 1's
There were 5 2's
There were 4 3's
There were no 4's
There were 3 5's
There were 5 6's
There were 4 7's
There was only 1 8 
There were 3 9's

这是我的C程序

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int i, j, array[30]={0}, count=0,check;
    srand(time(NULL));
    for(i=0;i<30;i++)
        array[i]=rand()%10;
    for(i=0;i<30;i++)
        printf("%d ",array[i]);
    printf("\n\n");

    for(i=29;i>=0;i--)
        printf("%d ",array[i]);
    printf("\n\n");

    for(i=0;i<30;i++){
        check=array[i];
        if(array[i]!=-1)
            array[i]=-1;

        if(check == -1)
            continue;

        count =1;

        for(j=0;j<30;j++){
            if((i==j) || (array[j]==-1))
                continue;

            if(check==array[j]){
                count++;
                array[j]=-1;
            }
        }

        printf("There were %d %d's\n",count,check);
    }

    return 0;
}

Answer 1

您将从注释中了解算法：

#include <stdio.h>
#include <stdlib.h>
//time.h is needed for time()
#include <time.h>

int main()
{
    int i, array[30] = {0};
    srand(time(NULL));

    //generate and print 30 random numbers
    for(i = 0; i < 30; i++){
        array[i] = rand() % 10;
        printf("%d ", array[i]);
    }

    puts("\n\n");

    //print these numbers backwards
    for(i = 29; i >= 0; i--)
        printf("%d ",array[i]);

    puts("\n\n");

    // print out a count of how many times each number
    // between 0 and 9 appeared in the sequence.
    int count[10] = {0};

    for(i = 0; i < 30; i++)
        count[array[i]]++;

    //output the count for each number
    for(i = 0; i < 10; i++)
        printf("There were %d %d's\n",count[i], i);

    return 0;
}

输出：

9 2 3 9 8 4 3 8 1 3 6 4 3 2 5 3 2 3 0 1 9 0 3 5 1 3 3 8 2 0

0 2 8 3 3 1 5 3 0 9 1 0 3 2 3 5 2 3 4 6 3 1 8 3 4 8 9 3 2 9

There were 3 0's
There were 3 1's
There were 4 2's
There were 9 3's
There were 2 4's
There were 2 5's
There were 1 6's
There were 0 7's
There were 3 8's
There were 3 9's

Answer 2

以下是在删除if语句时简化原始来源的信息。在逻辑位置用作源代码语句一部分的某些地方，隐含if语句。

例如，for(j = 0; j < 30 && match >= 0; j++)有几个逻辑表达式，但是此语句中没有if出现。逻辑表达式为j < 30和match >= 0，以及j < 30 && match >= 0的完整表达式。

此示例在语句array[j] == match && ++count && (array[j] = -1);中使用逻辑表达式和C编译器的评估短路行为（请参阅Wikipedia中的Short-circuit evaluation），以便逻辑表达式array[j] == match的计算结果为如果为false，则该语句的其余部分将不会执行。

我们还依靠带有++count的preincrement运算符来增加计数，然后采用结果值来检查它是否为假（零）或真（非零）。由于变量count初始化为零，并且递增时将始终为非零，因此该语句中的下一个逻辑表达式将被计算为(array[j] = -1)。我们将赋值语句放在括号内以强制执行求值顺序。我们希望为变量array[j]分配值-1，然后在逻辑语句中使用结果。因为这是整个逻辑语句的最后一个逻辑表达式，所以它求值为假（零）还是真（非零）都没有关系，因为我们想要的是将-1赋值给数组的副作用元素。

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

#include <time.h>

int main()
{
    int i, array[30] = {0};

    srand(time(NULL));
    for(i = 0; i < 30; i++)
        array[i] = rand() % 10;

    for(i = 0; i < 30; i++)
        printf("%d ", array[i]);
    printf("\n\n");

    for(i = 29; i >= 0; i--)
        printf("%d ",array[i]);
    printf("\n\n");

    for(i = 0; i < 30; i++){
        int j;
        int count = 0;
        int match = array[i];
        for(j = 0; j < 30 && match >= 0; j++){
            array[j] == match && ++count && (array[j] = -1);
            // replaces the following if as value of count is
            // tested after it is incremented so will always be nonzero.
            //   if (array[j] == match) {
            //       count++; array[j] = -1;
            //   }
        }

        // if this is a valid array element value we are trying to match
        // then print the count and the value being matched. printf()
        // is a function that returns an int indicating number of character written.
        match >= 0 && printf("There were %d %d's\n", count, match);
    }

    return 0;
}

示例输出。

8 2 4 0 8 0 8 1 1 4 6 9 3 9 7 6 3 9 0 1 0 7 1 2 4 0 3 0 2 3

3 2 0 3 0 4 2 1 7 0 1 0 9 3 6 7 9 3 9 6 4 1 1 8 0 8 0 4 2 8

There were 3 8's
There were 3 2's
There were 3 4's
There were 6 0's
There were 4 1's
There were 2 6's
There were 3 9's
There were 4 3's
There were 2 7's

C编程：替换if语句

2 个答案: