c – 编译器在编译时如何检测编号溢出？

编译器将源代码作为字符串处理,所以在C中,例如当它鼓励语句如unsigned char x = 150;它从类型限制中知道unsigned char必须在0到255之间.

我的问题是数字150仍然是字符串,编译器使用什么算法来比较数字序列 – 在这种情况下为150 – 对类型限制？

我为一个十进制,八进制,十六进制和小端二进制的类型’int’做了一个简单的算法,但我不认为编译器会这样做以检测数字溢出.

我所做的算法用C编码：

typedef signed char int8;
typedef signed int  int32;

#define DEC  0
#define HEX  1
#define OCT  2
#define BIN  3

bool isOverflow(const char* value,int32 basE)
{
    // left-most digit for maximum and minimum number
    static const char* max_numbers[4][2] =
    {
        //                 INT_MAX                           INT_MIN
        {                       "2147483647","2147483648" },// decimal
        {                         "7fffffff","80000000" },// hexadecimal
        {                      "17777777777","20000000000" },// octal
        { "01111111111111111111111111111111","10000000000000000000000000000000" }  // binary
    };

    // size of Strings in max_numbers array
    static const int32 number_sizes[] = { 10,8,11,32 };

    // input String size
    int32 str_len = strlen(value);

    // is sign mark exist in input String
    int32 signExist = ((base == DEC || base == OCT) && *value == '-');

    // first non zero digit in input number
    int32 non_zero_index = signExist;

    // locate first non zero index
    while(non_zero_index < str_len && value[non_zero_index] == 0) non_zero_index++;

    // if non_zero_index equal length then all digits are zero
    if (non_zero_index == str_len) return false;

    // get number of digits that actually represent the number
    int32 diff = str_len - non_zero_index;

    // if Difference less than 10 digits then no overflow will happened
    if (diff < number_sizes[base]) return false;
    // if Difference greater than 10 digits then overflow will happened
    if (diff > number_sizes[base]) return true;

    // left digit in input and search Strings
    int8 left1 = 0,left2 = 0;

    // if Digits equal to 10 then loop over digits from left to right and compare
    for (int32 i = 0; non_zero_index < str_len; non_zero_index++,i++)
    {
        // get input digit
        left1 = value[non_zero_index];
        // get match digit
        left2 = max_numbers[signExist][i];

        // if Digits not equal then if left1 is greater overflow will occurred,false otherwise
        if (left1 != left2) return left1 > left2;
    }

    // overflow won't happened
    return false;
}

可以优化该算法以适用于所有整数类型,但是使用浮点数我必须使用新的算法来处理IEEE浮点表示.

我认为编译器使用有效的算法来检测除了我的溢出,不是吗？