This question already has answers here:
Closed 10 years ago.
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how do I print an unsigned char as hex in c++ using ostream?
Convert ASCII string into Decimal and Hexadecimal Representations
I want to print out the hex value of characters using isprint(). However, I cannot get it to work. This is my attempt:
getline(cin, w);
for(unsigned int c = 0; c < w.size(); c++)
{
if(!isprint(w[c]))
{
cout << "ERROR: expected <value> found " << hex << w[c] << endl;
return 0;
}
}
Can anyone help me print out this hex value? Thanks! I'm inputting things like:
í
and I want it's hex value.
By default, a char is printed as an string character. Try casting the char to a general int like this:
cout << "ERROR: expected <value> found " << hex << static_cast<int>(w[c]) << endl;
Related
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String plus Char - what is happening?
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C++. Why std::cout << char + int prints int value?
(2 answers)
cout and String concatenation
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Closed 10 months ago.
I was experimenting with a statement in C++ using online compilers. When I try to run this specific code
cout << num[i] + " " + num[i];
The online compilers give no output. I can change the + symbol to << but I want to know the reason that the code does not give any output on these online compilers.
Online compilers that I tried are onlinegdb, programiz, and jdoodle.
#include <iostream>
#include <string>
int main() {
std::string num = "123";
int i = 0;
std::cout << num[i] + " " + num[i];
return 0;
}
C++ is not like JavaScript or many higher-level languages, as in you may not delimit you data with +'s or ,'s. As shown in Lewis' answer, each item you wish to have printed must be separated by an insertion delimiter (<<). As for extracting, you may use the extraction delimiter (>>).
In your case, you are doing mathematical operations on the the characters themselves (adding together their numerical ASCII representations together, which could print unprintable and invisible characters). The printable ASCII characters range from 32 (space character) to 127 (delete character) (base 10). When summing '1' + ' ' + '1' you are left with (49 + 32 + 49) or (130) which exceeds the printable character range. Or you may also be accessing garbage as #pm100 said in the comments due to pointer arithmetic.
Here is an example of using the insertion operator:
#include <iostream>
int main(void) {
int some_int = 1;
std::cout << "this is my " << some_int << "st answer on StackOverflow :)"
<< std::endl;
return 0;
}
And as for the extraction operator:
#include <iostream>
int main(void) {
int num;
std::cout << "Enter an integer: ";
std::cin >> num; // stores the input into the `num` variable
std::cout << "The number is: " << num << std::endl;
return 0;
}
Pointer arithmetic:
const char* get_filename(const char* _path, size_t _offset) {
return (_path + _offset);
}
// This is an example
//
// path = "path/to/my/file/file.txt";
// offset ^ ^
// 0 |
// + 16 ------------|
// path = "file.txt";
I am working on below code:
#include<iostream>
#include<stdio.h>
using namespace std;
main() {
unsigned char a;
a=1;
printf("%d", a);
cout<<a;
}
It is printing 1 and some garbage.
Why cout is behaving so?
cout << a is printing a value which appears to be garbage to you. It is not garbage actually. It is just a non-printable ASCII character which is getting printed anyway. Note that ASCII character corresponding to 1 is non-printable. You can check whether a is printable or not using, std::isprint as:
std::cout << std::isprint(a) << std::endl;
It will print 0 (read: false) indicating the character is non-printable
--
Anyway, if you want your cout to print 1 also, then cast a to this:
cout << static_cast<unsigned>(a) << std::endl;
I had a similar issue here that I've long forgotten about. The resolution to this problem with iostream's cout can be done like this:
#include<iostream>
#include<stdio.h>
main() {
unsigned char a;
a=1;
printf("%d", a);
std::cout<< +a << std::endl;
return 0;
}
instead of casting it back to another type if you want cout to print the unsigned char value as opposed to the ascii character. You need to promote it.
If you noticed all I did was add a + before the unsigned char. This is unary addition that will promote the unsigned char to give you the actual number representation.
User Baum mit Augen is responsible for reminding me of this solution.
You need to typecast a as integer as cout<< (int)(a);. With this you will observe 1 on the output. With cout << a;, the print will be SOH (Start of Heading) corresponding to ascii value of 1 which can't be printed and hence, some special character is observed.
EDIT:
To be more accurate, the cout statement should be cout << static_cast<unsigned>(a) as Nawaz has mentioned.
The C compiler has its own way of defining the type of the printed output, because you can specify the type of the output.
Ex:
uint8_t c = 100;
printf("%d",c);
so you can also print c as an int by %d, or char %c, string %s or a hex value %x.
Where C++ has its own way too, the cout prints the 8-bit values as a char by default. So, you have to use specifiers with the output argument.
You can either use:
a + before the name of the output argument
uint8_t data_byte = 100;
cout << "val: " << +data_byte << endl;
use a function cast unsigned(var); like,
uint8_t data_byte = 100;
cout << "val: " << unsigned(data_byte) << endl;
printf("%u",a);
its so simple try it
This question already has answers here:
C++ cout hex format
(3 answers)
Closed 9 years ago.
I'm trying to output a hex value assigned to a variable x and I can't seem to get it working in C++. I can do it in standard C but am getting undesired results in C++.
#include <iostream>
#include <iomanip>
using namespace std;
int main(int argc, const char * argv[])
{
unsigned char x = 0xFF;
printf("%X\n", x);
cout << dec << x << endl;
cout << hex << x << endl;
return 0;
}
prints
FF
\377
\377
Because it's unsigned char, the stream thinks you want to output a character, rather than its value. Try casting to int
cout << hex << (int)x << endl;
You might also want to use setw(2) and setfill('0') stream modifiers to pad single-digit hex numbers to 2 digits (similar to using %02X with printf).
I am writing a hex to dec conversion function. The input is a single character which is converted to hex and returned back as a char.
Here is the function
char hex2dec(char inp)
{
char out;
cout << "inp:" << inp;
if(inp >= '0' && inp <='9')
{
out = (inp - '0');
cout << " out " << out;
}
else
{
out = (toupper(inp) - 'A' + 10);
cout << " out " << out;
}
return out;
}
When i pass '0' and 'A' to the function, the print i get is
inp:0 out
inp:A out
i.e nothing is printed in out.
I am not able to find the issue..Can anyone help?
What you are actually trying to print is ascii characters with codes [0-15] which are not printable characters ie you want to print 15 but you print "\x0f" etc
use:
cout << (int)out;
and you'll force cout to invoke method printing ints not chars - this will solve your issue.
..or more 'c++++ ish' ;-)
cout << static_cast(out);
..or this which for most looks weird:
cout << int(out);
Use int out instead of char out. cout prints char as a character not an integer. The ASCII values 0-15 are unprintable control characters.
the reason why u got an 'A' from print is that out is char ,and after 'A' - 'A' + 10, out is a character whose ascii value is 10 instead of integer 10. So u will get a character whose ascii value is 10 instead of getting 'A'.After checking the ascii table, that character is null,which can explain why u get nothing in the output.
Simply use std::stringstream for this purpose.
#include <iostream>
#include <sstream>
int main (int argc, char ** argv)
{
// use hex formatting
std::stringstream stream;
stream << std::hex << 'A';
// retrieve value of the hex input character
int value;
stream >> value;
std::cout << value << std::endl; // prints 10
// reset the stream
stream.str ();
stream.clear();
// also works for strings
stream << "0xABC" << std::endl;
stream >> value;
std::cout << value << std::endl; // prints 2748
return 0;
}
I am working on below code:
#include<iostream>
#include<stdio.h>
using namespace std;
main() {
unsigned char a;
a=1;
printf("%d", a);
cout<<a;
}
It is printing 1 and some garbage.
Why cout is behaving so?
cout << a is printing a value which appears to be garbage to you. It is not garbage actually. It is just a non-printable ASCII character which is getting printed anyway. Note that ASCII character corresponding to 1 is non-printable. You can check whether a is printable or not using, std::isprint as:
std::cout << std::isprint(a) << std::endl;
It will print 0 (read: false) indicating the character is non-printable
--
Anyway, if you want your cout to print 1 also, then cast a to this:
cout << static_cast<unsigned>(a) << std::endl;
I had a similar issue here that I've long forgotten about. The resolution to this problem with iostream's cout can be done like this:
#include<iostream>
#include<stdio.h>
main() {
unsigned char a;
a=1;
printf("%d", a);
std::cout<< +a << std::endl;
return 0;
}
instead of casting it back to another type if you want cout to print the unsigned char value as opposed to the ascii character. You need to promote it.
If you noticed all I did was add a + before the unsigned char. This is unary addition that will promote the unsigned char to give you the actual number representation.
User Baum mit Augen is responsible for reminding me of this solution.
You need to typecast a as integer as cout<< (int)(a);. With this you will observe 1 on the output. With cout << a;, the print will be SOH (Start of Heading) corresponding to ascii value of 1 which can't be printed and hence, some special character is observed.
EDIT:
To be more accurate, the cout statement should be cout << static_cast<unsigned>(a) as Nawaz has mentioned.
The C compiler has its own way of defining the type of the printed output, because you can specify the type of the output.
Ex:
uint8_t c = 100;
printf("%d",c);
so you can also print c as an int by %d, or char %c, string %s or a hex value %x.
Where C++ has its own way too, the cout prints the 8-bit values as a char by default. So, you have to use specifiers with the output argument.
You can either use:
a + before the name of the output argument
uint8_t data_byte = 100;
cout << "val: " << +data_byte << endl;
use a function cast unsigned(var); like,
uint8_t data_byte = 100;
cout << "val: " << unsigned(data_byte) << endl;
printf("%u",a);
its so simple try it