函数原型

#include <stdlib.h>
int atoi(char * digital) 

#include <string.h>
char *strcpy(char *dest, const char *src);
void *memmove(void *dest, const void *src, size_t n);
char *strstr(const char *haystack, const char *needle);

具体实现：


int myAtoi(string str) {
    if (str.empty()) return 0;
    int neg = 1, base = 0, i = 0, n = str.size();
    while (i < n && str[i] == ' ')
        i++;
    if ((i < n) && (str[i] == '+' || str[i] == '-') ) {
        neg = (str[i] == '-') ? -1 : 1;
        i++;
    }
    for (; (i < n)&&( str[i] >= '0' && str[i] <= '9'); i++) {
        if (base > 0x7FFFFFFF / 10 || base == 0x7FFFFFFF / 10 && str[i] > '7') {
            return neg==-1 ? 0x80000000 : 0x7fffffff;
        }
        base = base * 10 + (str[i] - '0');
    }
    return base * neg;
}

char *strstr(const char *haystack, const char *needle) {
    char *cp = (char *) str1;
    char *s1, *s2; 
    if ( !*str2 ) 
        return((char *)str1); 
    while (*cp) { 
        s1 = cp; 
        s2 = (char *) str2; 
        while ( *s1 && *s2 && !(*s1-*s2) ) 
        s1++, s2++; 
        
        if (!*s2) return(cp); 
        cp++; 
    }
     return(NULL);
}

int strStr(string haystack, string needle) {
        if(haystack.empty() || needle.empty()) return 0;        
        int len = haystack.size();
        int len2= needle.size();

        int i=0;
        while(i < len) {            
            int j=0;
            while(j<len2){
                if(haystack[i+j] !=needle[j]) {
                    break;
                }
                j++;
            }
            if(j==len2) {
                return i;
            }
            i++;
        }
        return -1;
    }

char * strcpy(char * dest, const char *scr) {
    if(!dest || !scr) return nullptr;
    char *ret = dest ;
        int i = 0 ;
        for( ; src[i] != '\0'; i++) 
      dest[i] = src[i];
    }
    dest[i]='\0';
    return ret;
}

void *memmove(void *dest, const void *src, size_t n)
{
    if(!dest || !scr || n <=0 ) return nullptr;
    if(dest > scr ) {
        while(n--) {
           *(char*)(dest+n-1) = *(char*)(scr+n-1);
        }

    } else {
        while(n-->0) {
            *(char*)dest++ = *(char*)scr++;  
        }
    }
}