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.
Related
I am fairly new to C++ and have been tinkering with some simple programs to teach myself the basics. I remember a little while ago installing MinGW on my desktop (I think for the C compiler). I decided to go for Cygwin for C++ and it seems to have been working a treat, up until I noticed that it seems to behave differently from the MinGW compiler for this program. I'm probably breaking some coding golden rule, but the reason for this is that Windows CMD uses the MinGW compiler, or I can open the Cygwin shell and use that instead. Variety!
For the program below, I am making notes on the ternary operator and switch statement. I initialize 'a' and 'b', ask for user input, and then use a function to check which is larger.
I then ask for user input again, to over-write the values in 'a' and 'b' to something else. When compiling from Windows CMD, this works fine and I can overwrite with new input, and the #define MAXIMUM and MINIMUM functions work fine on the new values. When I compile on Cygwin, however, after the first 'pause', the program launches past the two std::cin's and just runs MAXIMUM and MINIMUM on the old values of 'a' and 'b'.
Obviously I have tried just creating two new int variables, 'c' and 'd', and then there is no issue. The values are not immutable in any sense that I am aware of (although I don't know much, so I could be wrong).
I wasn't sure if this was something to do with the auto keyword, so I specified the type as int manually.
I also checked the version of both compilers with 'g++ -v' in Cygwin and at the CMD.
#include <iostream>
#include <cstdlib>
#define MAXIMUM(a,b) ((a > b) ? a : b)
#define MINIMUM(a,b) ((a < b) ? a : b)
int ternary_op(int a, int b)
{
char x = 'a';
char y = 'b';
auto result_var = 0;
result_var = a > b ? x : y; //Shorthand if statement with syntax (test_condition) ? (if_true) : (if_false)
return result_var;
}
int main()
{
auto a = 0;
auto b = 0;
auto larger = 0;
auto smaller = 0;
std::cout << "Enter an integer: " << "\n";
std::cin >> a;
std::cout << "Enter another integer: " << "\n";
std::cin >> b;
char greater_var = ternary_op(a,b); //Therefore if condition a > b is satisfied, greater_var is assigned x ('a')
std::cout << greater_var << std::endl;
switch(greater_var){
case 'a':
std::cout << "First integer " << a << " is larger than second, " << b << std::endl;
break;
case 'b':
std::cout << "Second integer " << b << " is larger than first integer, " << a << std::endl;
break;
}
std::cout << system("cmd /c pause") << std::endl;
std::cout << "We can also use defined functions to check equivalency and assign variables based upon the result." << "\n";
std::cout << "Enter an integer: " << std::endl;
std::cin >> a;
std::cout << "Enter another integer: " << std::endl;
std::cin >> b;
larger = MAXIMUM(a,b);
smaller = MINIMUM(a,b);
std::cout << "Larger and smaller numbers determined by defined function: " << larger << ", " << smaller << std::endl;
std::cout << system("cmd /c pause") << std::endl;
return 0;
}
Obviously if I make two new variables, 'c' and 'd', there is no issue. Changing the type to int myself did not change the way the program behaved using Cygwin. Unusually the MinGW version was 8.1.0, while Cygwin was 7.4.0. I'm not sure if this means that it is simply an older version of the same compiler.
Again, I'm very new to this so I'm just quite confused as to why they would behave so differently. I was also under the impression that different compilers were completely different beasts that simply read from the same standard hymn sheet, so to speak.
Just curious as to what is going on here!
Cheers!
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 want to read a chunk of data from file into stringstream, which later will be used to parse the data (using getline, >>, etc). After reading the bytes, I set the buffer of the stringstream, but I cant make it to set the p pointer.
I tested the code on some online services, such as onlinegdb.com and cppreference.com and it works. However, on microsoft, I get an error - the pointers get out of order.
Here's the code, I replaced the file-read with a char array.
#include <sstream>
#include <iostream>
int main()
{
char* a = new char [30];
for (int i=0;i<30;i++)
a[i]='-';
std::stringstream os;
std::cout << "g " << os.tellg() << " p " << os.tellp() << std::endl;
os.rdbuf()->pubsetbuf(a,30);
os.seekp(7);
std::cout << "g " << os.tellg() << " p " << os.tellp() << std::endl;
}
the output I get when it works
g 0 p 0
g 0 p 7
the output I get on visual studio 2015
g 0 p 0
g -1 p -1
any ides?
thanks
std::sstream::setbuf may do nothing:
If s is a null pointer and n is zero, this function has no effect.
Otherwise, the effect is implementation-defined: some implementations do nothing, while some implementations clear the std::string member currently used as the buffer and begin using the user-supplied character array of size n, whose first element is pointed to by s, as the buffer and the input/output character sequence.
You are better off using the std::stringstream constructor to set the data or call str():
#include <sstream>
#include <iostream>
int main()
{
std::string str( 30, '-' );
std::stringstream os;
std::cout << "g " << os.tellg() << " p " << os.tellp() << std::endl;
os.str( str );
os.seekp(7);
std::cout << "g " << os.tellg() << " p " << os.tellp() << std::endl;
}
#include <conio.h> // include conio.h file
#include <iostream.h> // or #include<iostream>
int main()
{
int cout = 5;
cout << cout;
return 0;
}
Why does this happen ??
The code compiles correctly but it does not give expected output when it runs
This does not give the output 5 and all other stuff
It also does not give a warning.
The following line declares an int that happens to be called cout (it is not the std::cout stream)
int cout = 5;
The << operator peforms a bit shift.
So
cout << cout;
is only performing a bit shift and not storing the result.
To clarify, have a look at the following program:
#include<iostream>
int main()
{
int cout = 5;
auto shiftedval = cout << cout;
std::cout << "cout's value is " << cout << ", and the result of the bit shift is " << shiftedval << '\n';
return 0;
}
It will output:
cout's value is 5, and the result of the bit shift is 160
What is happening behind the scenes is that operator<< has been overloaded to take an ostream on the left hand side.
By including iostream you get this function and the compiler will know what you mean if you have an ostream to the left of the << operator.
Without a library the << would just simply have been a bitwise shift operator.
Also note that if you had ill-advisedly included using namespace std; or using std::cout then cout would then mean ostream and << would trigger a call to the library operator<< function. If after adding the using declaration above you include another declaration of cout the newly declared name will hide the previous declaration and cout will now be considered an int again and we're back to the bit shift operator functionality being used.
Example:
#include<iostream>
using namespace std; // using std:: at global scope
int main()
{
int cout = 5;
auto shiftedval = cout << cout;
//the following will not compile, cout is an int:
cout << "cout's value is " << cout << ", and the result of the bit shift is " << shiftedval << '\n';
//but we can get the cout from the global scope and the following will compile
::cout << "cout's value is " << cout << ", and the result of the bit shift is " << shiftedval << '\n';
return 0;
}
You are naming your variable cout, which you are confusing with std::cout. Your example preforms a bit shift operation. Try this instead :
int main()
{
int cout = 5;
std::cout << cout;
return 0;
}
Better yet, name your variable anything else to avoid the confusion :
int main()
{
int foo = 5;
std::cout << foo;
return 0;
}
If you don't explicitly declare the std namespace, either by including using namespace std; in your code or calling std::cout then cout resolves to the local variable cout in your main() function.
Even if you do declare using namespace std; cout will still resolve to the local variable instead - this is one reason why a lot of people, books, and tutorials will recommend that you explicitly write std::whatever instead of using the namespace.
Why after using strdup(value) (int)value returns you different output than before?
How to get the same output?
My short example went bad, please use the long one:
Here the full code for tests:
#include <stdio.h>
#include <iostream>
int main()
{
//The First Part
char *c = "ARD-642564";
char *ca = "ARD-642564";
std::cout << c << std::endl;
std::cout << ca << std::endl;
//c and ca are equal
std::cout << (int)c << std::endl;
std::cout << (int)ca << std::endl;
//The Second Part
c = strdup("ARD-642564");
ca = strdup("ARD-642564");
std::cout << c << std::endl;
std::cout << ca << std::endl;
//c and ca are NOT equal Why?
std::cout << (int)c << std::endl;
std::cout << (int)ca << std::endl;
int x;
std::cin >> x;
}
Because an array decays to a pointer in your case, you are printing a pointer (ie, on non-exotic computers, a memory address). There is no guarantee that a pointer fits in an int.
In the first part of your code, c and ca don't have to be equal. Your compiler performs a sort of memory optimization (see here for a full answer).
In the second part, strdup allocates dynamically a string twice, such that the returned pointers are not equal. The compiler does not optimize these calls because he does not seem to control the definition of strdup.
In both cases, c and ca may not be equal.
"The strdup() function shall return a pointer to a new string, which is a duplicate of the string pointed to by s1." source
So it's quite understandable that the pointers differ.