How to check if char * p has reached the end of line C?

Question

template<class IntType> IntType atoi_unsafe(const char* source) { IntType result = IntType(); while (source) { auto t = *source; result *= 10; result += (*source - 48); ++source; } return result; } 

and in main() I have:

 char* number = "14256"; atoi_unsafe<unsigned>(number); 

but the while (source) condition does not seem to recognize that source has iterated over the entire string C. How to check the end of the string correctly?

Answer 1

while(source) true until the pointer breaks to 0, but most likely will work long before that on modern systems. You need to dereference the pointer to find the null byte, while(*source) .

_{I hate posting short answers}

Answer 2

The pointer does not end at the end of the line; end of line is found when the specified value becomes zero. Hence:

 while (*source != '\0') 

You can, more compactly, write the entire function as follows:

 template<class IntType> IntType atoi_unsafe(const char* source) { IntType result = IntType(); char c; while ((c = *source++) != '\0') result = result * 10 + (c - '0'); return result; } 

Of course, it does not use the auto keyword. Also note the difference between '\0' and '0' . Brackets in assignment in the loop body are not needed.

Your code only processes unsigned strings - and perhaps confirms that the characters are actually numbers (perhaps throwing an exception if the input is invalid). An “unsafe” appeal certainly applies. Note also that if you instantiate a template for a signed integer type and overflow the value, you invoke undefined behavior. At least with unsigned types, arithmetic is defined, even if probably not what is expected.

Answer 3

You need to search for a null terminator at the end of the line. Waiting for the pointer to wrap around 0 will probably never happen. Use while (*source) for your loop.