你可以自己卷!

#include <stdio.h>
#include <string.h>
#include <math.h>

int my_atoi(const char* snum)
{
    int idx, strIdx = 0, accum = 0, numIsNeg = 0;
    const unsigned int NUMLEN = (int)strlen(snum);

    /* Check if negative number and flag it. */
    if(snum[0] == 0x2d)
        numIsNeg = 1;

    for(idx = NUMLEN - 1; idx >= 0; idx--)
    {
        /* Only process numbers from 0 through 9. */
        if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39)
            accum += (snum[strIdx] - 0x30) * pow(10, idx);

        strIdx++;
    }

    /* Check flag to see if originally passed -ve number and convert result if so. */
    if(!numIsNeg)
        return accum;
    else
        return accum * -1;
}

int main()
{
    /* Tests... */
    printf("Returned number is: %d\n", my_atoi("34574"));
    printf("Returned number is: %d\n", my_atoi("-23"));

    return 0;
}

这样就可以做你想做的事情，而且不会杂乱。

这个函数会帮助你

int strtoint_n(char* str, int n)
{
    int sign = 1;
    int place = 1;
    int ret = 0;

    int i;
    for (i = n-1; i >= 0; i--, place *= 10)
    {
        int c = str[i];
        switch (c)
        {
            case '-':
                if (i == 0) sign = -1;
                else return -1;
                break;
            default:
                if (c >= '0' && c <= '9')   ret += (c - '0') * place;
                else return -1;
        }
    }

    return sign * ret;
}

int strtoint(char* str)
{
    char* temp = str;
    int n = 0;
    while (*temp != '\0')
    {
        n++;
        temp++;
    }
    return strtoint_n(str, n);
}

裁判:http://amscata.blogspot.com/2013/09/strnumstr - - 2.版本的html

在c++中，你可以使用这样一个函数:

template <typename T>
T to(const std::string & s)
{
    std::istringstream stm(s);
    T result;
    stm >> result;

    if(stm.tellg() != s.size())
        throw error;

    return result;
}

这可以帮助您将任何字符串转换为任何类型，如float, int, double…

int atoi(const char* str){
    int num = 0;
    int i = 0;
    bool isNegetive = false;
    if(str[i] == '-'){
        isNegetive = true;
        i++;
    }
    while (str[i] && (str[i] >= '0' && str[i] <= '9')){
        num = num * 10 + (str[i] - '0');
        i++;
    }
    if(isNegetive) num = -1 * num;
    return num;
}

//I think this way we could go :
int my_atoi(const char* snum)
{
 int nInt(0);
 int index(0);
 while(snum[index])
 {
    if(!nInt)
        nInt= ( (int) snum[index]) - 48;
    else
    {
        nInt = (nInt *= 10) + ((int) snum[index] - 48);
    }
    index++;
 }
 return(nInt);
}

int main()
{
    printf("Returned number is: %d\n", my_atoi("676987"));
    return 0;
}

好吧，我也有同样的问题。我想出了这个解决方案。这对我来说是最有效的。我试过用atoi()，但效果不太好。下面是我的解决方案:

void splitInput(int arr[], int sizeArr, char num[])
{
    for(int i = 0; i < sizeArr; i++)
        // We are subtracting 48 because the numbers in ASCII starts at 48.
        arr[i] = (int)num[i] - 48;
}