# Convert char to int in C and C++

## The Question :

447 people think this question is useful

How do I convert a char to an int in C and C++?

• @Matt: it would be a good idea to be more concrete. asking about a generalization just invites generalized answers that are not applicable or even correct for your task. keep in mind, when you have to ask, you probably don’t know enough to generalize correctly.
• @Alf P. Steinbach: The original question was vague regarding which language. With keywords c and c++, I think answers confronting both languages are reasonable.
• From my extensive experience on other technical forums, my intuition is that the OP really means “how do I take the textual representation of a number (in base 10) and convert it to the corresponding number?” Generally speaking, C and C++ neophytes usually have incredibly fuzzy ideas about how text works in those languages and what char really means.
• @KarlKnechtel: If that’s true (I give it about 50/50 as lots of early tutorials also encourage getting ASCII values out of chars, even though ASCII doesn’t cover the full range), the OP needs to clarity – but that’s a dupe of stackoverflow.com/questions/439573/….
• The OP had three hours to clarify this question and failed to do so. As it is, there’s no way to know what is actually asked. Voted to close.

605 people think this answer is useful

Depends on what you want to do:

to read the value as an ascii code, you can write

char a = 'a';
int ia = (int)a;
/* note that the int cast is not necessary -- int ia = a would suffice */



to convert the character '0' -> 0, '1' -> 1, etc, you can write

char a = '4';
int ia = a - '0';
/* check here if ia is bounded by 0 and 9 */



Explanation:
a - '0' is equivalent to ((int)a) - ((int)'0'), which means the ascii values of the characters are subtracted from each other. Since 0 comes directly before 1 in the ascii table (and so on until 9), the difference between the two gives the number that the character a represents.

115 people think this answer is useful

Well, in ASCII code, the numbers (digits) start from 48. All you need to do is:

int x = (int)character - 48;



Or, since the character ‘0’ has the ASCII code of 48, you can just write:

int x = character - '0';  // The (int) cast is not necessary.



62 people think this answer is useful

C and C++ always promote types to at least int. Furthermore character literals are of type int in C and char in C++.

You can convert a char type simply by assigning to an int.

char c = 'a'; // narrowing on C
int a = c;



34 people think this answer is useful

char is just a 1 byte integer. There is nothing magic with the char type! Just as you can assign a short to an int, or an int to a long, you can assign a char to an int.

Yes, the name of the primitive data type happens to be “char”, which insinuates that it should only contain characters. But in reality, “char” is just a poor name choise to confuse everyone who tries to learn the language. A better name for it is int8_t, and you can use that name instead, if your compiler follows the latest C standard.

Though of course you should use the char type when doing string handling, because the index of the classic ASCII table fits in 1 byte. You could however do string handling with regular ints as well, although there is no practical reason in the real world why you would ever want to do that. For example, the following code will work perfectly:

  int str[] = {'h', 'e', 'l', 'l', 'o', '\0' };

for(i=0; i<6; i++)
{
printf("%c", str[i]);
}



You have to realize that characters and strings are just numbers, like everything else in the computer. When you write ‘a’ in the source code, it is pre-processed into the number 97, which is an integer constant.

So if you write an expression like

char ch = '5';
ch = ch - '0';



this is actually equivalent to

char ch = (int)53;
ch = ch - (int)48;



which is then going through the C language integer promotions

ch = (int)ch - (int)48;



and then truncated to a char to fit the result type

ch = (char)( (int)ch - (int)48 );



There’s a lot of subtle things like this going on between the lines, where char is implicitly treated as an int.

18 people think this answer is useful

(This answer addresses the C++ side of things, but the sign extension problem exists in C too.)

Handling all three char types (signed, unsigned, and char) is more delicate than it first appears. Values in the range 0 to SCHAR_MAX (which is 127 for an 8-bit char) are easy:

char c = somevalue;
signed char sc = c;
unsigned char uc = c;
int n = c;



But, when somevalue is outside of that range, only going through unsigned char gives you consistent results for the “same” char values in all three types:

char c = somevalue;
signed char sc = c;
unsigned char uc = c;
// Might not be true: int(c) == int(sc) and int(c) == int(uc).
int nc = (unsigned char)c;
int nsc = (unsigned char)sc;
int nuc = (unsigned char)uc;
// Always true: nc == nsc and nc == nuc.



This is important when using functions from ctype.h, such as isupper or toupper, because of sign extension:

char c = negative_char;  // Assuming CHAR_MIN < 0.
int n = c;
bool b = isupper(n);  // Undefined behavior.



Note the conversion through int is implicit; this has the same UB:

char c = negative_char;
bool b = isupper(c);



To fix this, go through unsigned char, which is easily done by wrapping ctype.h functions through safe_ctype:

template<int (&amp;F)(int)>
int safe_ctype(unsigned char c) { return F(c); }

//...
char c = CHAR_MIN;
bool b = safe_ctype<isupper>(c);  // No UB.

std::string s = "value that may contain negative chars; e.g. user input";
std::transform(s.begin(), s.end(), s.begin(), &amp;safe_ctype<toupper>);
// Must wrap toupper to eliminate UB in this case, you can't cast
// to unsigned char because the function is called inside transform.



This works because any function taking any of the three char types can also take the other two char types. It leads to two functions which can handle any of the types:

int ord(char c) { return (unsigned char)c; }
char chr(int n) {
assert(0 <= n);  // Or other error-/sanity-checking.
assert(n <= UCHAR_MAX);
return (unsigned char)n;
}

// Ord and chr are named to match similar functions in other languages
// and libraries.



ord(c) always gives you a non-negative value – even when passed a negative char or negative signed char – and chr takes any value ord produces and gives back the exact same char.

In practice, I would probably just cast through unsigned char instead of using these, but they do succinctly wrap the cast, provide a convenient place to add error checking for int-to-char, and would be shorter and more clear when you need to use them several times in close proximity.

13 people think this answer is useful

Use static_cast<int>:

int num = static_cast<int>(letter); // if letter='a', num=97



Edit: You probably should try to avoid to use (int)

int num = (int) letter;

7 people think this answer is useful

It sort of depends on what you mean by “convert”.

If you have a series of characters that represents an integer, like “123456”, then there are two typical ways to do that in C: Use a special-purpose conversion like atoi() or strtol(), or the general-purpose sscanf(). C++ (which is really a different language masquerading as an upgrade) adds a third, stringstreams.

If you mean you want the exact bit pattern in one of your int variables to be treated as a char, that’s easier. In C the different integer types are really more of a state of mind than actual separate “types”. Just start using it where chars are asked for, and you should be OK. You might need an explicit conversion to make the compiler quit whining on occasion, but all that should do is drop any extra bits past 256.

7 people think this answer is useful

I have absolutely null skills in C, but for a simple parsing:

char* something = "123456";

int number = parseInt(something);



…this worked for me:

int parseInt(char* chars)
{
int sum = 0;
int len = strlen(chars);
for (int x = 0; x < len; x++)
{
int n = chars[len - (x + 1)] - '0';
sum = sum + powInt(n, x);
}
return sum;
}

int powInt(int x, int y)
{
for (int i = 0; i < y; i++)
{
x *= 10;
}
return x;
}



3 people think this answer is useful

Presumably you want this conversion for using functions from the C standard library.

In that case, do (C++ syntax)

typedef unsigned char UChar;

char myCppFunc( char c )
{
return char( someCFunc( UChar( c ) ) );
}



The expression UChar( c ) converts to unsigned char in order to get rid of negative values, which, except for EOF, are not supported by the C functions.

Then the result of that expression is used as actual argument for an int formal argument. Where you get automatic promotion to int. You can alternatively write that last step explicitly, like int( UChar( c ) ), but personally I find that too verbose.

Cheers & hth.,

0 people think this answer is useful

I was having problems converting a char array like "7c7c7d7d7d7d7c7c7c7d7d7d7d7c7c7c7c7c7c7d7d7c7c7c7c7d7c7d7d7d7c7c2e2e2e" into its actual integer value that would be able to be represented by 7C’ as one hexadecimal value. So, after cruising for help I created this, and thought it would be cool to share.

This separates the char string into its right integers, and may be helpful to more people than just me 😉

unsigned int* char2int(char *a, int len)
{
int i,u;
unsigned int *val = malloc(len*sizeof(unsigned long));

for(i=0,u=0;i<len;i++){
if(i%2==0){
if(a[i] <= 57)
val[u] = (a[i]-50)<<4;
else
val[u] = (a[i]-55)<<4;
}
else{
if(a[i] <= 57)
val[u] += (a[i]-50);
else
val[u] += (a[i]-55);
u++;
}
}
return val;
}



Hope it helps!

0 people think this answer is useful

For char or short to int, you just need to assign the value.

char ch = 16;
int in = ch;



Same to int64.

long long lo = ch;



All values will be 16.

int charToint(char a){
`