char变量是否总是用{'x'，\\ 0}表示为已取消引用的char * [2]？

Question

I tried this method to convert a char digit to an integer digit ('3'->3) and it seems to work.

char c='x';
int i=atoi(&c);

My question is, will this always work?

Is there always a NULL character after the first character?

There is another way to do this using implicit typecast, but I am not sure it is a good practice. Actually, I am pretty surprised there are no warnings even using -Wall and -Wextra.

int i = c - '0';

Note I am using GCC 4.8.2 and MinGW.

Answer 1

This compiles but it's undefined behavior.

A char is just a character. No null byte following. std::atoi expects a null byte, though, so invoking std::atoi(&c) compiles but yields undefined behavior. This line,

int i = std::atoi("x");

is well-defined, though, because the string "x" is a char[2] and null-terminated.

int i = c - '0';

is alright because ¹ the C++11 standard and the C99 standard ² guarantee the order of character digits.

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_{¹ as @ShafikYaghmour noted in the comments to your question}

_{² Thanks to @ShafikYaghmour again!}

Answer 2

atoi expects null terminated C-string.

since you didn't provide it (you provided a pointer to character) this is undefined behaviour.

for the second line of code - this generally works if you can guarantee that the character is indeed a digit. for example , the character # will yield a result but may not the result you'd expect.

some solution is to use std::stoi :

try{ int x = std::stoi(std::string() + c) } catch(...) {/*handle*/}

Answer 3

This will work as long as next to the memory space where c is allocated is a \\0. In other words, it's just luck. You should use int i = c - '0'

But be aware that every character will be mapped to int (fe int i = 'a' - '0' )

To avoid this you could do something like:

int i = c=>'0' && c<='9' ? c-'0' : -1; //-1 represents not valid values