C language: a daily exercise (selection + programming)

Table of contents

 Multiple choice questions:

Question one:

Question two:

Question three: 

Question four: 

Question five:

Programming questions:

Question 1: Print 1 to the largest n digits

Example 1

Idea one:

Question 2: Calculate the date to day conversion

Example 1

 Idea one:

My limited strength may not explain and understand some places clearly enough, you can try to read the code yourself, or point out mistakes in the comment area, hope Haihan!

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 Multiple choice questions:

Question one:

1. Execute the following program, the correct output is ( )
int x=5,y=7;
void swap()
{         int z;         z=x;         x=y;         y=z; }

int main()
{
        int x=3,y=8;
        swap();
        printf("%d,%d\n"，x, y);
        return 0;
}

 A: 5,7         B: 7,5         C: 3,8         D: 8,3

Question two:

2. The following incorrect definition statement is ( )
A: double x[5] = {2.0, 4.0, 6.0, 8.0, 10.0};
B: char c2[] = {'\x10', '\xa', ' \8'};
C: char c1[] = {'1','2','3','4','5'}; D:
int y[5+3]={0, 1, 3 , 5, 7, 9};

Question three: 

3. The test.c file includes the following statement. Among the four variables defined in the file, the variable of pointer type is [multiple choices] ( ) 

#define INT_PTR int*
typedef int* int_ptr;
INT_PTR a, b;
int_ptr c, d;

 A: a         B: b         C: c         D: d

Question four: 

4. If the conditional expression (M)?(a++):(a--) is given, the expression M ( )
A: and (M==0) are equivalent to B: and (M==1), etc. price

C: Equivalent to (M!=0)          D: Equivalent to (M!=1)

Question five:

5. If there is the following definition statement, the correct input statement is [multiple choice] ( )
int b;
char c[10];

A: scanf("%d%s",&b,&c);         B: scanf("%d%s",&b,c);
C: scanf("%d%s",b,c);              D: scanf("%d%s",b,&c); 

Programming questions:

Question 1: Print 1 to the largest n digits

Example 1

enter:

1

return value:

[1,2,3,4,5,6,7,8,9]

Idea one:

        Create a pointer array a , use num to record the number of elements that need to be input , input the values ​​from 1 to num to each array subscript, and return the pointer a .

Note: returnSize here refers to the number of array elements!

int* printNumbers(int n, int* returnSize ) 
{
    int i = 0;
    int num = 1;
    int* a;
    //计算元素个数
    for(i = n;i != 0;i--)
    {
        num *= 10; 
    }
    *returnSize =num-1;
    //开辟需要存储的个数的空间
    a = (int*)malloc(sizeof(int)*(*returnSize));
    for(i = 1;i < num;i++)
    {
        a[i-1] = i;
    }
    return a;
}

Question 2: Calculate the date to day conversion

Example 1

enter:

2012 12 31

output:

366

 Idea one:

        Create year, mon, day for scanf() , and month[] to record the number of days in each month , then judge whether the year is a leap year, and then sum the number of days.

#include <stdio.h>

int main() 
{
    int year = 0;
    //保存每月天数
    int month[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
    int day = 0;
    int mon = 0;
    int sum = 0;
    scanf("%d%d%d",&year,&mon,&day);
    //判断是否是闰年
    if((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
    {
        month[2] = 29;
    }
    //求目标月份前的天数和
    for(int i = 1;i < mon ;i++)
    {
        sum += month[i];
    }
    printf("%d",sum+day);


    return 0;
}

My limited strength may not explain and understand some places clearly enough, you can try to read the code yourself, or point out mistakes in the comment area, hope Haihan!