C Language—Daily Multiple Choice Questions—Day53

Pointer related blogs

The first shot of pointer: pointer family-CSDN Blog

In-depth understanding: dereference and addition of pointer variables-CSDN Blog

first question

1. There is the following program, the output result is ()

#include <stdio.h>
int main() 
{
    char a = 'H';
    a = (a > 'A' && a <= 'Z') ? (a + 32) : a;
    printf("%c\n", a);
    return 0;
}

A：H

B：h

C：A

D：a

Answer and analysis B

This question tests the understanding of ASCII values ​​and conditional expressions

Conditional expression: x? y:z; if x is true, execute y, otherwise execute z;

Obviously this question is true, the execution is a + 32, which is the operation of converting uppercase characters to lowercase letters

Below is the ASCII table:

Question 2

2. Which of the four variables defined is not a pointer type?

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

A：a

B：b

C：c

D：d

E; are all pointers

F: None of them are pointers

Answer and analysis B

Remember #define is replacement, so

INT_PTR a,b = int* a, b, replacement means only one is given, not both by default;

Typedef is a renaming of the type, so int_ptr is int*, both pointer types;

Question 3

3. The result of running the following program is ()

#include <stdio.h>
int f(int n)
{
  if (n==1)
    return 1;
  else
    return (f(n-1)+n*n*n);
}
int main()
{
  int s=f(3);
  printf("%d\n", s);
  return 0;
}

A：8

B：9

C：27

D：36

Answer and analysisD

This question examines a simple function recursion

f(3) = f(2) + 3 * 3 * 3

f(2) = f(1) + 2 * 2 * 2

f(1) = 1

so

f(3) = f(1) + 2 * 2 * 2 + 3 * 3 * 3 = 1 + 2 ^ 3 + 3 ^ 3 = 36

Question 4

4. What will the following code output ()

#include <stdio.h>
int main()
{
    int a[4] = {1, 2, 3, 4};
    int *ptr = (int *)(&a + 1);
    printf("%d", *(ptr - 1));
}

A：1

B：2

C：3

D：4

Answer and analysisD

This question tests the dereference and addition of pointers. Please see this blog: In-depth understanding: Dereference and addition of pointer variables - CSDN Blog

Question 5

5. The output of the following program is ()

#include <stdio.h>
int main() 
{
    int a[12] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, *p[4], i;
    for (i = 0; i < 4; i++) 
        p[i] = &a[i * 3];
    printf("%d\n", p[3][2]);
    return 0;
}

A: There is an error in the above program

B：6

C：8

D：12

Answer and analysisD

This question is very interesting. First, there is an integer array with 12 variables, and then a pointer array is defined. This pointer array stores &a[0], &a[3], &a[6], &a[9 respectively. ]These four addresses;

When printing, it is p[3][2]. How to understand this?

Use pointers to understand: p[3] [2] = *(p[3] + 2) This means getting the element p[3] first, which is &a[9]

So p[3] [2] = *(&a[9] + 2); that is, the length of 2 int types after the address of a[9], why is it int?

Look at the fourth question blog, and then you come to &a[11], so in the end it is *&a[11] = a[11] = 12;

In-depth understanding: dereference and addition of pointer variables-CSDN Blog