Question

我正在尝试此代码将整数转换为字符串它正在打印值，但未将任何内容返回给调用函数

char* itoa(int num, int num_len)
{
    char str[num_len+2];
    int i;
    //if the num is 0
    if(num == 0)
        strcpy(str, "0");
    //if the num is negetive then we append a '-' sign at the beginning
    //and put '\0' at (num_len+1)th position 
    else if(num < 0)
    {
        num *= -1;
        str[num_len+1] = '\0';
        str[0] = '-';
        i = num_len+1;
    }
    //we put '\0' (num_len)th position i.e before the last position
    else
    {
        str[num_len] = '\0';
        i = num_len;
    } 

    for(;num>0;num/=10,i--)
    {
        str[i] = num%10 + '0';
        printf("%c ",str[i]);//for debugging
    }

    return str;
}

Answer 1

我只是忘记了那个规则。非常感谢你

char* itoa(int num, int num_len,char* str)
{
    int i;
    str = (char*)malloc((num_len + 2)*sizeof(char)); 
    if(num == 0)
        strcpy(str, "0");
    else if(num < 0)
    {
        num *= -1;
        str[num_len+1] = '\0';
        str[0] = '-';
        i = num_len;
    }
    else
    {
        str[num_len] = '\0';
        i = num_len-1;
    } 

    for(;num>0;num/=10,i--)
    {
        str[i] = num%10 + '0';
        printf("%c ",str[i]);
    }

    return str;
}

我正在尝试制作一个将整数转换为不返回字符串的字符串的函数。我该如何解决？

1 个答案: