题一：
模拟实现strncpy
#include <stdio.h>
void my_strncpy(char* arr2, char* arr1, size_t num)
{
int i = 0;
for (i = 0; i < num; i++)
{
*(arr2 + i) = *(arr1 + i);
}
}
int main()
{
char arr1[] = "hello liangzai";
char arr2[10] = { 0 };
//strncpy(arr2,arr1,sizeof(arr2));
my_strncpy(arr2, arr1, sizeof(arr2));
return 0;
}
优解一：
char * mystrncpy(char * dst, const char * src, size_t n)
{
int i;
for (i = 0; src[i] && i < n; i++)
{
dst[i] = src[i];
}
if (i < n)
{
dst[i] = 0;
}
return dst;
}
题二：
模拟实现strncat
#include <stdio.h>
#include <string.h>
char* my_strncat(char* arr1, const char* arr2, size_t num)
{
int sz = strlen(arr1);
int i = 0;
for (i = 0; i < num; i++)
{
*(arr1 + sz + i) = *(arr2 + i);
}
return arr1;
}
int main()
{
char arr2[] = "shawanyi ";
char arr1[20] = "hello ";
//strncat(arr1,arr2,10);
my_strncat(arr1, arr2, 10);
return 0;
}
优解一：
char * mystrncat(char * dst, const char * src, size_t n)
{
char * tmp = dst;
while (*dst)
{
dst++;//记录到dst最后一个位置
}
int i;
for (i = 0; src[i] && i < n; i++)
{
dst[i] = src[i];
}
dst[i] = 0;
return tmp;
}
题三：
模拟实现atoi
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <ctype.h>
int my_atoi(const char* arr1)
{
assert(arr1);
int sum = 0;
if (*(arr1) == '\0')
{
return 0;
}
while(isspace(*arr1))
{
arr1++;
}
int flat = 1;
if (*arr1 == '+')
{
arr1++;
}
else if (*arr1 == '-')
{
flat = -1;
arr1++;
}
int ret = 0;
while (isdigit(*arr1))
{
ret = ret * 10 + (*arr1 - '0');
if (ret > INT_MAX)
{
return INT_MAX * flat;
}
else if (ret < INT_MIN)
{
return INT_MIN * flat;
}
arr1++;
}
if (isdigit(*arr1) == 0)
{
return ret * flat;
}
return ret*flat;
}
int main()
{
char arr1[] = "-123666a44";
//int n = atoi(arr1);
int sz = 0;
sz = my_atoi(arr1);
return 0;
}
优解一
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
//VALID 表示结果合法
//INVALID 表示结果非法
enum State
{
VALID,//0
INVALID//1
};
//默认结果可能非法，当正确转换后再改为合法
enum State state = INVALID;
int my_atoi(char *str)
{
int flag = 1;
long long ret = 0;
assert(str);
state = INVALID;
//跳过空白字符
while(isspace(*str))
{
str++;
}
if(*str == '\0')
{
return 0;
}
//跳过正负号
if(*str == '+')
{
str++;
}
else if(*str == '-')
{
flag = -1;
str++;
}
//开始转换数字字符直到非数字字符
while(isdigit(*str))
{
ret = ret * 10 + flag * (*str-'0');
if((ret > INT_MAX) || (ret < INT_MIN))
{
return 0;
}
str++;
}
//正常停止
if(*str == '\0')
{
state = VALID;
return (int)ret;
}
else
{
//遇到非数字字符
return (int)ret;
}
