When should I use single quotes and double quotes in C or C++ programming?
In C and in C++ single quotes identify a single character, while double quotes create a string literal. 'a' is a single a character literal, while "a" is a string literal containing an 'a' and a null terminator (that is a 2 char array).
In C++ the type of a character literal is char, but note that in C, the type of a character literal is int, that is sizeof 'a' is 4 in an architecture where ints are 32bit (and CHAR_BIT is 8), while sizeof(char) is 1 everywhere.
Some compilers also implement an extension, that allows multi-character constants. The C99 standard says:
6.4.4.4p10: "The value of an integer character constant containing more
than one character (e.g., 'ab'), or
containing a character or escape
sequence that does not map to a
single-byte execution character, is
implementation-defined."
This could look like this, for instance:
const uint32_t png_ihdr = 'IHDR';
The resulting constant (in GCC, which implements this) has the value you get by taking each character and shifting it up, so that 'I' ends up in the most significant bits of the 32-bit value. Obviously, you shouldn't rely on this if you are writing platform independent code.
Single quotes are characters (char), double quotes are null-terminated strings (char *).
char c = 'x';
char *s = "Hello World";
'x' is an integer, representing the numerical value of the
letter x in the machine’s character set
"x" is an array of characters, two characters long,
consisting of ‘x’ followed by ‘\0’
I was poking around stuff like: int cc = 'cc'; It happens that it's basically a byte-wise copy to an integer. Hence the way to look at it is that 'cc' which is basically 2 c's are copied to lower 2 bytes of the integer cc. If you are looking for a trivia, then
printf("%d %d", 'c', 'cc'); would give:
99 25443
that's because 25443 = 99 + 256*99
So 'cc' is a multi-character constant and not a string.
Cheers
Single quotes are for a single character. Double quotes are for a string (array of characters). You can use single quotes to build up a string one character at a time, if you like.
char myChar = 'A';
char myString[] = "Hello Mum";
char myOtherString[] = { 'H','e','l','l','o','\0' };
single quote is for character;
double quote is for string.
In C, single-quotes such as 'a' indicate character constants whereas "a" is an array of characters, always terminated with the \0 character
Double quotes are for string literals, e.g.:
char str[] = "Hello world";
Single quotes are for single character literals, e.g.:
char c = 'x';
EDIT As David stated in another answer, the type of a character literal is int.
A single quote is used for character, while double quotes are used for strings.
For example...
printf("%c \n",'a');
printf("%s","Hello World");
Output
a
Hello World
If you used these in vice versa case and used a single quote for string and double quotes for a character, this will be the result:
printf("%c \n","a");
printf("%s",'Hello World');
output :
For the first line. You will get a garbage value or unexpected value or you may get an output like this:
�
While for the second statement, you will see nothing. One more thing, if you have more statements after this, they will also give you no result.
Note: PHP language gives you the flexibility to use single and double-quotes easily.
Use single quote with single char as:
char ch = 'a';
here 'a' is a char constant and is equal to the ASCII value of char a.
Use double quote with strings as:
char str[] = "foo";
here "foo" is a string literal.
Its okay to use "a" but its not okay to use 'foo'
Single quotes are denoting a char, double denote a string.
In Java, it is also the same.
While I'm sure this doesn't answer what the original asker asked, in case you end up here looking for single quote in literal integers like I have...
C++14 added the ability to add single quotes (') in the middle of number literals to add some visual grouping to the numbers.
constexpr int oneBillion = 1'000'000'000;
constexpr int binary = 0b1010'0101;
constexpr int hex = 0x12'34'5678;
constexpr double pi = 3.1415926535'8979323846'2643383279'5028841971'6939937510;
In C & C++ single quotes is known as a character ('a') whereas double quotes is know as a string ("Hello"). The difference is that a character can store anything but only one alphabet/number etc. A string can store anything.
But also remember that there is a difference between '1' and 1.
If you type
cout<<'1'<<endl<<1;
The output would be the same, but not in this case:
cout<<int('1')<<endl<<int(1);
This time the first line would be 48. As when you convert a character to an int it converts to its ascii and the ascii for '1' is 48.
Same, if you do:
string s="Hi";
s+=48; //This will add "1" to the string
s+="1"; This will also add "1" to the string
different way to declare a char / string
char char_simple = 'a'; // bytes 1 : -128 to 127 or 0 to 255
signed char char_signed = 'a'; // bytes 1: -128 to 127
unsigned char char_u = 'a'; // bytes 2: 0 to 255
// double quote is for string.
char string_simple[] = "myString";
char string_simple_2[] = {'m', 'S', 't', 'r', 'i', 'n', 'g'};
char string_fixed_size[8] = "myString";
char *string_pointer = "myString";
char string_poionter_2 = *"myString";
printf("char = %ld\n", sizeof(char_simple));
printf("char_signed = %ld\n", sizeof(char_signed));
printf("char_u = %ld\n", sizeof(char_u));
printf("string_simple[] = %ld\n", sizeof(string_simple));
printf("string_simple_2[] = %ld\n", sizeof(string_simple_2));
printf("string_fixed_size[8] = %ld\n", sizeof(string_fixed_size));
printf("*string_pointer = %ld\n", sizeof(string_pointer));
printf("string_poionter_2 = %ld\n", sizeof(string_poionter_2));
Say I wanted to shift the letter 'A' into the letter 'D'. I can't seem to find anything that shows how to do that in C++ 17. Any suggestions?
Try just adding 3 to the character:
char myChar = 'A';
char shifted = myChar + 3; // shifted is now 'D'
Just treat each character like an integer and shift ‘A’ according to its ASCII value. This works because in c++ characters are encoded as 7-bit integers.
http://www.asciitable.com
Looking at the table we see that ‘A’ = 65 and ‘D’ = 68. So to shift ‘A’ to ‘D’, simply add 3 to ‘A’ like so:
char a = 'A';
a += 3;
std::cout << a;
Output:
D
char letter_A='A';
char letter_D=letter_A+3;
In the following code, I can not understand why the string is converted to int in this way.
Why is it using a sum with 0 ?
string mystring;
vector<int> myint;
mystring[i+1]=myint[i]+'0';
This code converts an int (presumably a digit) to the character that represents it.
Since characters are sequential, and chars can be treated as integers, the character representing a certain digit can, in fact, be described by its distance from '0'. This way, 0 turns turn to the character '0', '5' is the character that is greater than '0' by five, and so on.
This is an efficient, old school and dangerous method to get a char representation of a single digit. '0' will be converted to an int containing its ASCII code (0x30 for '0') and then that is added to myint[i]. If myint[i] is 9 or lower, you can cast myint[i] to a char you will get the resulting digit as text.
Things will not go as expected if you add more than 9 to '0'
You can also get a number from its char representation :
char text = '5';
int digit = text - '0';
The '0' expression isn't string type, it's char type that stores characters of ASCII and also can represent numbers from 0 to 255. So, in arithmetic operations char behaves like integer type.
In C strings a represent as arrays of char: static (char str[N]) or dynamic (char *str = new char[n]). String literals puts into double quotes ("string").
So, '0' is char and "0" is char[1]
Reading through c++ document . I found char or unsigned char stores a byte.
However as in below code. If I wish to store 1 or any numbers or characters as it is.
char c = 1;
std::cout << c; // it gives smiley symbol.
std::cout << int(c); // now gives desired output.
Now if I have a string and want to append char to string , I do,
std::string bla = "hello";
bla.push_back(c); // this appends the smiley symbol not 1.
bla.push_back(int(c)); // still appends the smiley symbol.
bla.push_back(int(c -'0')); // appends D.
I wish to append exact 8 bit as it is in a string . Any suggestion
1 is a character literal in the ASCII table. 1 is not the same as '1'. Wrap the char around quotes to print out 1. char c = '1';
You missed the quote. Make the declaration char c = '1';
It will solve your problem.
I have const binary data that I need insert to buffer
for example
char buf[] = "1232\0x1";
but how can do it when binary data is at first like below
char buf[] = "\0x11232";
compiler see it like a big hex number
but my perpose is
char buf[] = {0x1,'1','2','3','2'};
You can use compile-time string concatenation:
char buf[] = "\x01" "1232";
However, with a 2-digit number after \x it also works without:
char buf[] = "\x011232";
You can create a single string literal by composing it of adjacent strings - the compiler will concatenate them:
char buf[] = "\x1" "1232";
is equivalent to:
char buf[] = {0x1,'1','2','3','2', 0}; // note the terminating null, which may or may not be important to you
You have to write it in two byte or four byte format:
\xhh = ASCII character in hexadecimal notation
\xhhhh = Unicode character in hexadecimal notation if this escape sequence is used in a wide-character constant or a Unicode string literal.
so in your case you have to write "\x0112345"