1. Simulate the implementation of strncat

Among the system library functions, there is the strncat function, which is used for appending strings, that is, appending a string after a string. Its function prototype is:

char *strncat(char *strDest,const *strSource,size_t count);

*strDest is the target string *strSource is the source string, count is the number of strings to be appended.

1.1 Code writing

#include<stdio.h>
#include<assert.h>
#include<stdlib.h>
typedef unsigned int uint;
char *my_strncat(char *dest, const char *src, uint count)//模拟实现strncat函数
{
 assert(dest);
 assert(src);
 int *ret = dest;
 while (*dest)
 {
  dest++; //找到dest中的\0
 }
 while (count--)
 {
  *dest++ = *src++;
 }
 *dest = '\0';
 return ret;
}
//程序测试
int main()
{
 char arr[20] = "abcdef";
 int len = strlen(arr);
 my_strncat(arr, arr, len);
 printf("%s\n", arr);
 system("pause");
 return 0;
}

1.2 The results are as follows

2. Simulate strncpy

char *strncpy(char *strDest,const *strSource,size_t count);

*strDest is the target string *strSource is the source string, count is the number of characters that need to be copied from the source string to the target string.

Note: If the length of the source string is less than count, add 0 to the target string after copying the source string, until the count is

2.1 Code writing

char * My_strncpy(char * dest, const char *src, size_t n)
{
    assert(dest);
    assert(src);

    char *ret = dest;
    while (n--)
    {
        *dest++ = *src++;
    }
    return ret;
}
int main()
{
    char str1[20] = "123456789";
    char str2[20] = "abcde";
    printf("%s", My_strncpy(str1, str2, 3));

}

2.2 The results are as follows

3. Find a single dog

Only two numbers in an array appear once, all other numbers appear twice.

Write a function to find these two numbers that occur only once.

3.1 Code writing

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#include<assert.h>
void find_two_diff_num(int arr[], int sz, int *p1, int *p2)
{
	int i = 0;//循环变量
	int ret = 0;
	int pos = 0;
	*p1 = 0;//数字1的地址
	*p2 = 0;//数字2 的地址
	//1.把所有数字异或 
	for (i = 0; i < sz; i++)
	{
		ret ^= arr[i];//循环到最后一次的结果是5^6,即就是101^110=011
	}
	//2.找ret二进制中为1的一位
	for (i = 0; i < 32; i++)
	{
		if (((ret >> i) & 1) == 1)//上一步两个数异或的后两位为1
		{
			pos = i;
			break;
		}
	}
	//分组
	for (i = 0; i < sz; i++)
	{
		if (((arr[i] >> pos) & 1) == 1)
		{
			(*p1) ^= arr[i];
		}
	}
	(*p2) = (*p1) ^ ret;
}


int main()
{
	int arr[] = { 1, 2, 5, 6, 8, 1, 6, 2, 9, 8 };
	int sz = sizeof(arr) / sizeof(arr[0]);//数组大小
	int num1 = 0;//数字1
	int num2 = 0;//数字2
	find_two_diff_num(arr, sz, &num1, &num2);
	printf("num1=%d,num2=%d\n", num1, num2);
	system("pause");
	return 0;
}

C language function simulation implementation