Operator exercises

 

Catalog

1 Knowledge point

2 Enter an integer n , Output the number 32 In bit binary representation 1 The number of .

3  Find the number of different bits in two binary numbers

4  Print odd and even bits of integer binary

5  The result of the following code is

6  Judge integer parity （ Multiple input scenarios ）

7 Judge whether it's a vowel or a consonant （ Multiple input scenarios ）

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1 Knowledge point

（1）b = ++c, c++, ++a, a++ // The comma expression has the lowest priority , Let's start here b=++c, b Get is ++c After the results of the ,b=++c And the following form a comma expression , Calculated from left to right .

（2） Global variables   When not initialized   It will be initialized to... By default 0.

（3） About expression evaluation , Be careful   Want to consider   Implicit type conversion 、 Arithmetic conversion 、 The priority of the operator   associativity   Order of evaluation

2 Enter an integer n , Output the number 32 In bit binary representation 1 The number of .

Negative numbers are represented by complements . requirement ： Write a function Returns the binary value of the parameter 1 The number of .

Code 1 Exhibition ：（ Be careful ： Here is the binary of complement in memory 1 The number of ）

#include <stdio.h>
int count_number_of_1(unsigned int n)
{
	int count = 0;
	while (n)
	{
		if (n % 2 == 1)
		{
			count++;
			n = n / 2;
		}
	}
	return count;
}
int main()
{
	int a =-1;
	int ret = count_number_of_1(a);
	printf("%d", ret);
	return 0;
}

Be careful ： The function says unsigned int （ Unsigned integer int)

Code parsing ：

7 The binary sequence of is 00000000000000000000000000000111, When 7%2=1, This 1 Is the last bit of binary ,7/2=3, This 3 yes 7 The binary sequence of moves one displayed digit to the right ,00000000000000000000000000000011.（ This line of thought, , When a decimal number , To know every digit , We usually go first %10, Divided by 10 Follow this cycle ）

Code 2 Exhibition ：

#include <stdio.h>
int count_number_of_1(int n)
{
	int count = 0;
	int i = 0;
	for (i = 0; i < 32; i++)
	{
		if ((n & 1) == 1)
		{
			count++;
		}
		n = n >> 1;
	}
	return count;
}

int main()
{
	int a = -1;
	int ret = count_number_of_1(a);
	printf("%d", ret);
	return 0;
}

Code 3 Exhibition :

#include <stdio.h>
int count_number_of_1(int n)
{
	int count = 0;
	while (n)
	{
		n = n & (n - 1);
		count++;
	}
	return count;
}

int main()
{
	int a = -1;
	int ret = count_number_of_1(a);
	printf("%d", ret);
	return 0;
}

Code understanding :

m&（m-1） Will remove one binary sequence from the right 1  （ 0111&0110=0110  0110&0101=0100）

（ Remove one at a time 1, How many times , Just how many 1）

( This is the best answer , There are several 1, Just cycle a few times .)

3  Find the number of different bits in two binary numbers

Two int（32 position ） Integers m and n In the binary representation of , How many bits (bit) Different

Code 1 Exhibition ：

#include <stdio.h>
int count_diff_bit(int m, int n)
{
	int count = 0;
	int i = 0;
	for (i = 0; i < 32; i++)
	{
		if ((m & 1) != (n & 1))
		{
			count++;
		}
		m >>= 1;
		n >>= 1;
	}
	return count;
}

int main()
{
	int a = -1;
	int b = 1;
	int ret = count_diff_bit(a, b);
	printf("%d", ret);
	return 0;
}

Code 2 Exhibition ：

#include <stdio.h>
int count_diff_bit(int m, int n)
{
	int count = 0;
	int i = m ^ n;
	while (i)
	{
		i = i & (i - 1);
		count++;
	}
	return count;
}

int main()
{
	int a = -1;
	int b = 1;
	int ret = count_diff_bit(a, b);
	printf("%d", ret);
	return 0;
}

  Exclusive or Same as 0, Different for 1., The result goes back to finding a binary sequence 1 The number of .

4  Print odd and even bits of integer binary

Get all even and odd bits in an integer binary sequence , Print out the binary sequence separately

#include <stdio.h>
void print(int m)
{
	int i = 0;
	for (i = 30; i >= 0; i -= 2)
	{
		printf("%d ", (m >> i) & 1);
	}

	printf("\n");
	for (i = 31; i >= 1; i -= 2)
	{
		printf("%d ", (m >> i) & 1);
	}
}

int main()
{
	int a = 5;
	scanf("%d", &a);
	print(a);
	return 0;
}

5  The result of the following code is

#include <stdio.h>
int i;
int main()
{
    i--;
    if (i > sizeof(i))
    {
        printf(">\n");
    }
    else
    {
        printf("<\n");
    }
    return 0; 
}

A.>         B.<       C. No output         D. There is a problem with the procedure

  analysis ：

C In language ,0 For false , Not 0 It's true . Global variables , When no initial value is given , Compiling it initializes it to... By default 0.

i The initial value of 0,i-- result -1,i For shaping ,sizeof(i) seek i The type size is 4, According to this analysis , The result should be B, however sizeof The return value type of is actually an unsigned integer , Therefore, the compiler will automatically set the left i Automatically convert to unsigned data ,-1 The corresponding unsigned integer is a very large number , exceed 4 perhaps 8, Therefore, it should be selected in practice A

sizeof  Operands   The result type of calculation is size_t  Unsigned integer unsigned int    In this question  int >unsigned int  First of all, we should put int type （ Perform arithmetic conversion in implicit type conversion ） convert to  unsigned int  Type is OK   When -1  When it is considered an unsigned integer （-1 The complement of is 32 individual 1）, Original code   Inverse code   The complement should be the same , that -1 The unsigned integer of is obviously very large , Greater than 4, So we should choose A

6  Judge integer parity （ Multiple input scenarios ）

Enter any integer from the keyboard （ Range -231~231-1）, Program to judge its parity .      Input description ： Multi group input , Each line of input includes an integer .    Output description ： Enter... For each line , The output number is odd （Odd） Or even （Even）.

Code display ：

#include <stdio.h>
int main()
{
	int a = 0;
	while ((scanf("%d", &a)) != EOF)
	{
		if (a % 2 == 0)
		{
			printf("Even\n");
		}
		else
		{
			printf("Odd\n");
		}
	}
	return 0;
}

scanf Read failed , It will return EOF

7 Judge whether it's a vowel or a consonant （ Multiple input scenarios ）

Input description ：

Multi group input , Enter one letter per line .

Output description ：

Enter... For each group , Output as a line , If the input letter is a vowel （ Include case ）, Output “Vowel”, If the input letter is a non vowel , Output “Consonant”.

#include <stdio.h>
int main()
{
	char arr[] = { 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U' };
	char ch = 0;
	while ((scanf("%c", &ch) != EOF))
	{
		int i = 0;
		for (i = 0; i < 10; i++)
		{
			if (ch == arr[i])
			{
				printf("Vowel\n");
				break;
			}
		}
		if (i == 10)
		{
			printf("Consonant\n");
		}
		getchar();
	}
}

Be careful ：

Get on the keyboard ch When it comes to type , Will be able to \n Also enter into , So want to use getchar（） Clean buffer

Code 2 Exhibition ：

#include <stdio.h>
int main()
{
	char arr[] = { 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U' };
	char ch = 0;
	while ((scanf("%c\n", &ch) != EOF))
	{
		int i = 0;
		for (i = 0; i < 10; i++)
		{
			if (ch == arr[i])
			{
				printf("Vowel\n");
				break;
			}
		}
		if (i == 10)
		{
			printf("Consonant\n");
		}
	}
}

Code 3 Exhibition ：

#include <stdio.h>
int main()
{
	char arr[] = { 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U' };
	char ch = 0;
	while ((scanf(" %c", &ch) != EOF))
	{
		int i = 0;
		for (i = 0; i < 10; i++)
		{
			if (ch == arr[i])
			{
				printf("Vowel\n");
				break;
			}
		}
		if (i == 10)
		{
			printf("Consonant\n");
		}
	}
}