I am trying to print these data type. But I get a very strange output instead of what I expect.
#include <iostream>
using namespace std;
int main(void)
{
char data1 = 0x11;
int data2 = 0XFFFFEEEE;
char data3 = 0x22;
short data4 = 0xABCD;
cout << data1 << endl;
cout << data2 << endl;
cout << data3 << endl;
cout << data4 << endl;
}
Most likely you expect data1 and data3 to be printed as some kind of numbers. However, the data type is character, which is why C++ (or C) would interpret them as characters, mapping 0x11 to the corresponding ASCII character (a control character), similar for 0x22 except some other character (see an ASCII table).
If you want to print those characters as number, you need to convert them to int prior to printing them out like so (works for C and C++):
cout << (int)data1 << endl;
Or more C++ style would be:
cout << static_cast<int>(data1) << endl;
If you want to display the numbers in hexadecimal, you need to change the default output base using the hex IO manipulator. Afterwards all output is done in hexadecimal. If you want to switch back to decimal output, use dec. See cppreference.com for details.
cout << hex << static_cast<int>(data1) << endl;
Related
I am trying to convert an integer to a char pointer as shown below. The data results are different. I am not sure what is going wrong. Please help me in correcting the code.
int main(){
char *key1 = "/introduction";
std::ostringstream str1;
str1<< 10;
std::string data=str1.str();
std::cout <<"The data value="<<data<<std::endl; // The data value= 10
char *intro= new char[data.length()+1];
strcpy(intro, data.c_str());
std::cout <<"The data value="<<*intro <<std::endl; // The data value=1
return 0;
}
I am not sure why two data value are printed different i.e, 10 and 1.
In C++, when trying to print all the contents of a char * with cout, you should pass the pointer, i.e. cout << intro << endl.
What you've done here is dereferenced the char *, so cout << *intro << endl is equivalent to cout << intro[0] << endl, which is equivalent to printing the first character, 1.
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
The code asks for a positive integer, than the first output shows the corresponding ASCII code, the others are made to convert the integer to decimal, octal and hexadecimal equivalents. I understand the logic of the code, but I don't understand the assignment made on line 10 c=code than the assignment made on line 12 code=c. What happens on background when we 'swap' the two variables.
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
unsigned char c = 0;
unsigned int code = 0;
cout << "\nPlease enter a decimal character code: ";
cin >> code;
c = code;
cout << "\nThe corresponding character: " << c << endl;
code = c;
cout << "\nCharacter codes"
<< "\n decimal: " << setw(3) << dec << code
<< "\n octal: " << setw(3) << oct << code
<< "\n hexadecimal: " << setw(3) << hex << code
<< endl;
return 0;
}
I could be wrong here so maybe someone else can weigh in, but I believe I know the answer.
If you assign a character a number, when you print that char it prints the corresponding character. Since c is of type char, the line c = code converts the integer entered into a character. You can test this yourself by assigning any int to a char variable and printing it out.
The second assignment, code = c, seems to be completely unnecessary.
That's not a swap. c is assigned the same value as code and then this value is assigned back to code. The original value of code is lost.
We can see this because unsigned char c is (usually) much smaller than unsigned int code and some information may be lost stuffing the value in code into c.
For example, code = 257. After c = code; code is still 257 and c, assuming an 8 bit char will be 1. After code = c;, both code and c will be 1. 257 has been lost.
Why is this being done? when given a char, operator<< will print out the character encoded, completely ignoring the request to print as hex, dec, or oct. So
<< "\n decimal: " << setw(3) << dec << c
is wasted. Given an int << will respect the modifiers, but if c and code have different values, you're comparing apples and Sasquatches.
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
I am learning C++ for the first time. I have no previous programming background.
In the book I have I saw this example.
#include <iostream>
using::cout;
using::endl;
int main()
{
int x = 5;
char y = char(x);
cout << x << endl;
cout << y << endl;
return 0;
}
The example makes sense: print an integer and the ASCII representation of it.
Now, I created a text file with these values.
48
49
50
51
55
56
75
I am writing a program to read this text file -- "theFile.txt" -- and want to convert these numbers to the ASCII value.
Here is the code I wrote.
#include <iostream>
#include <fstream>
using std::cout;
using std::endl;
using std::ifstream;
int main()
{
ifstream thestream;
thestream.open("theFile.txt");
char thecharacter;
while (thestream.get(thecharacter))
{
int theinteger = int(thecharacter);
char thechar = char(theinteger);
cout << theinteger << "\t" << thechar << endl;
}
system ("PAUSE");
return 0;
}
This is my understanding about the second program shown.
The compiler does not know the exact data type that is contained in "theFile.txt". As a result, I need to specify it so I choose to read the data as a char.
I read the each digit in the file as a char and converted it to an integer value and stored it in "theinteger".
Since I have an integer in "theinteger" I want to print it out as a character but char thechar = char(theinteger); does not work as intended.
What am I doing incorrect?
You are reading char by char, but you really (I think) want to read each sequence of digits as an integer. Change your loop to:
int theinteger;
while (thestream >> theinteger )
{
char thechar = char(theinteger);
cout << thechar << endl;
}
+1 For a very nicely formatted & expressed first question, BTW!
You are reading one char at a time from the file. Hence, if your file contains:
2424
You will first read the char "2" from the file, convert it to an int, and then back to a char, which will print "2" on cout. Next round will print "4", and so on.
If you want to read the numbers as full numbers, you need to do something like:
int theinteger;
thestream >> theinteger;
cout << char(theinteger) << endl;