广研二面题目

1. 实现一个memcpy函数。

先看下标准memcpy()的解释：

void *memcpy(void *dst, const void *src, size_t n);
//If copying takes place between objects that overlap, the behavior is undefined.

事实上所说的陷阱也在于此，自己动手实现memcpy()时就需要考虑地址重叠的情况。

另外，标准库也提供了地址重叠时的内存拷贝函数：memmove()，那么为什么还要考虑重写memcpy()函数呢?

因为memmove()函数的实现效率问题，该函数把源字符串拷贝到临时buf里，然后再从临时buf里写到目的地址，增加了一次不必要的开销。

void* Memcpy(void *dest, const void *src, size_t n)
{
    if(dest == NULL || src == NULL)
        return NULL;

    const char *psrc = (char *)src;
    char *pdest = (char *)dest;
    size_t i;

    if((psrc < pdest) && (psrc + n > pdest))
    {   
        i = n;
        while(i > 0) //注意i类型是size_t,无符号整形，所以不能用i>=0来判断
        {   
            pdest[i-1] = psrc[i-1]; 
            i--;
        }   
    }   
    else
    {   
        i = 0;
        while(i < n)
        {   
            pdest[i] = psrc[i]; i++
        }   
    }   
    return dest;
}

int main()
{
    char buf[100] = "abcdefghijk";
    Memcpy(buf+2, buf, 5); 
    printf("%s\n", buf);
    return 0;
}

2. 实现一个atoi的函数

double atof(char s[])
{
    double val = 0.0;
    int i,v_pow,sign,e_sign,e_pow;

    for (i = 0; isspace(s[i]); i++) // skip the white space
        ;   
    sign = (s[i] == '-') ? -1 : 1;
    if (s[i] == '-' || s[i] == '+')
        i++;
    for (val = 0; isdigit(s[i]); i++)
        val = val * 10 + (s[i] - '0');
    if (s[i] == '.')
        i++;
    for (v_pow = 1; isdigit(s[i]); i++)
    {   
        val = val * 10 + (s[i] - '0');
        v_pow *= 10; 
    }   
    if (s[i] == 'e' || s[i] == 'E') // handle scientific notation
        i++;
    e_sign = (s[i] == '-') ? -1 : 1;
    if (s[i] == '-' || s[i] == '+')
        i++;
    for (e_pow = 0; isdigit(s[i]); i++)
        e_pow = e_pow * 10 + (s[i] - '0');
    e_pow *= e_sign;
    return sign * (val * pow(10, (double)e_pow)) / v_pow;
}

double atoi(char s[])
{
    return (int)atof(s);
}

int main()
{
    char s[] = "-45.67e-100";
    double d = atof(s);
    printf("%g\n", d);
    char ss[] = "-45e4";
    int i = atoi(ss);
    printf("%d\n", i);
    return 0;
}