Consider the following code:
extern "C" {
#include <lib.h>
}
#include <iostream>
int main() {
unsigned char a='a';
unsigned char b=some_struct_in_libh->unsignedchar;
cout << a << " " << b << endl; //Prints only a
printf("%u\n",b); //Prints b
cout << static_cast<int>(b) << endl; //Also prints b
return 0;
}
Why does it behave like this?
It's not printing only a at all. What you're seeing is instead that cout prints character type data as characters not as numbers. Your b is some character that's non-printable so cout is helpfully printing it as a blank space.
You found the solution by casting it to int.
EDIT: I'm pretty sure your printf is only working by chance because you told it to expect an unsigned int and gave it a character (different number of bytes).
Related
#include <iostream>
using namespace std;
int main() {
int age = 20;
const char* pDept = "electronics";
cout << age << " " << pDept;
}
The above code is normal.
Why shouldn't I use cout << *pDept instead of cout << pDept above?
Both of them are legal in C++. Which one to use depends on what you want to print.
In your case, pDept is a pointer that points to a char in memory. It also can be used as a char[] terminated with \0. So std::cout << pDept; prints the string the pointer is pointing to.
*pDept is the content that pDept points to, which is the first character of the string. So std::cout << *pDept; prints the first character only.
I am using sscanf to put a MAC address from a string into a uint8 array. For some reason, the uint8 array is all blank.
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
std::string mac = "00:00:00:00:00:00";
uint8_t smac[7];
memset(smac, 0, 7);
sscanf(
mac.c_str(),
"%hhu:%hhu:%hhu:%hhu:%hhu:%hhu",
&smac[0],
&smac[1],
&smac[2],
&smac[3],
&smac[4],
&smac[5]
);
std::cout << "string: " << mac << std::endl;
std::cout << "uint8_t: "<< smac;
return 0;
}
uint8_t is on most platforms a typedef for unsigned char. Therefore, cout is trying to print it as a string, but it encounters a null byte (or string terminator) as the first character, so it stops printing.
A solution here would be to print all the MAC address members individually:
for(int c = 0; c < sizeof(smac); c++)
{
std::cout << +smac[c];
if(c != sizeof(smac) - 1)
std::cout << "::";
}
std::cout << '\n';
The + here performs integer promotion so smac[c] will be printed as a number and not a character.
The types uint8_t and unsigned char are generally equivalent to the compiler. The convention for outputting an array of char (unsigned or not) is to stop when you reach a value of zero, because that indicates the end of the string.
Hello guys i am beginner on the language c++
i was trying to run this code below on my ide"codeblocks" and it works
https://www.youtube.com/watch?v=vLnPwxZdW4Y (link for the tutorial that following )
#include <iostream>
using namespace std;
int main()
{
string charactername = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is" << charactername<< endl;
cout << "i am " << characterage << endl;
return 0;
}
this code does not work on my other compiler running on dosbox ? any ideas why ?
I suggest you stop using Turbo C++ as it is a very outdated and a discontinued compiler. However, if you don't have the option of using new compilers (I had the same issue as I studied C++ at school), you will have to make the following changes:
using namespace std; cannot be used in Turbo C++. You will have to remove that and replace #include<iostream> with #include<iostream.h>
Data-type string cannot be used in Turbo C++. You will have to declare a character array instead.
You will have to use #include<stdio.h> and the function puts(); to display the character array in case of Turbo C++. Alternatively you can use a loop-statement.
This will be your final code:
#include <iostream.h>
#include <stdio.h>
int main()
{
char charactername[] = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is ";
puts(charactername);
cout << "i am " << characterage << endl;
return 0;
}
Note: The puts(); function automatically puts the cursor on the next line. So you don't need to use endl;
Or, if you want to use a loop-statement to display the character array
#include <iostream.h>
int main()
{
char charactername[] = "arnold";
int characterage;
characterage = 10;
cout << "Hello my name is ";
int i=0;
while(charactername[i]!='\0') {
cout<<charactername[i];
i++;
}
cout<<endl;
cout << "i am " << characterage << endl;
return 0;
}
'\0' is the last element of the character array. So as long as the loop does not reach the last element, it will print the character array.
a[] = "arnold"; basically means an array is created like this: a[0]='a', a[1]='r', a[2]='n',.... a[5]='d', a[6]='\0'.
Alternatively, if you use cout << charactername; instead of the while loop, it will print the whole name. (This is only in the case of a string variable (character array), for an integer array or any other array you will need the while loop)
I am just playing around to see where my current understanding of C++ behaviour ends. I wrote the following code, and got some very unexpected results.
#include <iostream>
#include <cstddef>
using namespace std;
int main()
{
const char c = 'A';
cout << c << endl;
size_t l = (size_t)(&c);
char* d = (char*)l;
*d = 'B';
cout << (size_t)d << " " << (size_t)&c << endl;
cout << *d << " " << c << endl;
}
The first line outputs 'A', as expected. On the second line, the two addresses are the same. However, the third line outputs "B A".
Obviously this is terrible code, but why didn't the value of c change (or why didn't it fail to compile?) In other words, if they both have the same address, why don't they have the same value?
My system is GCC 4.8.1 64-bit on MS Windows 7, x86, if it matters.
I was experimenting with C++ and I decided to try the is_big_endian code, in much the same way I would do it in C. However I am getting no output when I try to print out the value of the pointer. I tried both the C and C++ style casts. What am I doing wrong?
#include <iostream>
using namespace std;
int main (void){
int num = 1;
char *ptr = (char *)#
//char *ptr = reinterpret_cast<char *>(&num);
cout << "Value is: " << *ptr << endl;
}
operator<< sees that you are outputting a char, so it prints it as a character, not as a number (it's as if in C you wrote %c instead of %d in a printf); and since *ptr will either be 0 or 1, you'll end up in both cases with a non-printable character.
To fix this, cast explicitly *ptr to int:
cout << "Value is: " << int(*ptr) << endl;