在Linux 32位上以36为基数的strtoull长时间溢出无符号

Question

Anyone know why using strtoull will overlow an unsigned long long? The value of "x" is 12 when I make the call, so I'm trying to convert a 12 digit base 36 num to unsigned long long.

This should work right? It doesn't matter whether I compile 32 or 64bit. I'm using g++ on redhat.

buffer is a char*

char *strPtr = buffer + ORDERIDOFFSET;
char *endPtr = strPtr + ORDERIDLENGTH;
long x = long((endPtr)) - long(buffer + ORDERIDOFFSET);
unsigned long long orderid = strtoull((buffer + ORDERIDOFFSET), &(endPtr), 36);

Thanks!

Answer 1

You seem to be under the misapprehension that endptr is an input parameter to strtoull , which tells it how long the input string is. It is not. It is an output parameter, which strtoull uses to tell you the first character it could not convert.

I presume that your buffer is not nul-terminated; probably there is an alphanumeric character following your 12-digit base-36 number.

By the way, imho, x would be better computed as endPtr - strPtr . Casting to long is not really correct.

Answer 2

You appear to be trying to use endPtr as an input parameter. std::strtoull takes the second parameter as an output parameter.

You may find it more robust to use std::stoull instead of std::strtoull :

char *strPtr = buffer + ORDERIDOFFSET;
char *endPtr = strPtr + ORDERIDLENGTH; // assuming ORDERIDLENGTH takes you 1 passed the end of the actual ID string
std::string s(strPtr, endPtr);
unsigned long long orderid = std::stoull(s, nullptr, 36);

Additionally,

long x = long((endPtr)) - long(buffer + ORDERIDOFFSET);

is not portable to 64-bit systems (you would truncate the address). I'm pretty sure you want

std::size_t x = endPtr - strPtr;