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#include <iostream> using namespace std; int main() { double u = 0; double w = -u; cout << w << endl; return 0; }
Why does this great piece of code outputs "-0" and not "0", as one would expect?
Started by Leonid on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
The IEEE 754 standard for floating point arithmetic....
In IEEE floating point 0 and -0 are both distinct values, from here under "Special Values":
Note that -0 and +0 are distinct values, though they both compare as equal.
Is zero.
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I'm running a program and redirecting cout to an outfile, like so:
./program < infile.in > outfile.o
I want to be able to read in an option '-h' or '--help' from the command line and output a help message to the terminal. Is there a way I can do...
Started by zebraman on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
Given....
Using shell redirection > only redirects stdout (cout redirection to outfile.o.
Std::cout << "This goes to stdout\n"; return 0; }
Will work like this:
$ ./a.out This goesYou should use cerr instead of cout.
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I don't get why I get 0 when I use printf and %d to get the size of my vector:
vector<long long> sieve; int size; ... //add stuff to vector ... size = sieve.size(); printf("printf sieve size: %d \n", size); //prints "printf sieve size: 0" std::cout...
Started by apphacker on
, 9 posts
by 9 people.
Answer Snippets (Read the full thread at stackoverflow):
Int main(){ std::cout << "sizeof(size_t): " << sizeof(size_t) << std::endl ; std::vector<long long> sieve ; std::cout << "sizeof(sieve.size()): " << sizeof())); return 0; } $ g++ trySize.C $ .....
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How can I derive a class from cout so that, for example, writing to it
new_cout << "message";
would be equivalent to
cout << __FUNCTION__ << "message" << "end of message" << endl;
Started by Jack on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
Class Log { public: Log(const std::string....
You may also create a new object or use existing objects like that .
")); return 0; } }
#define debug_print(message) (std::cout << __FUNCTION__ << (message to subclass std::cout.
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I want to print out a function pointer using cout, and found it did not work. But it worked after I converting the function pointer to (void *), so does printf with %p, such as
#include <iostream> using namespace std; int foo() {return 0;} int main...
Started by ibread on
, 6 posts
by 6 people.
Answer Snippets (Read the full thread at stackoverflow):
; ++first) { std::cout << std::hex << std::setw(2) << std::setfill('0') << // standard explicitly disallows this conversion cout << hex << fn; return 0; }
I note that my in that function'....
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Why is the output of the following code equals to 0 or serven?
cout << 7/9*9; //output 0 (zero) why? float nine = 9; float seven = 7; float i = seven/nine*nine; cout << i //output 7 Why?
Thanks for the help.
Started by Judy on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
When you made them:
7 / 9 = 0 0....
In:
cout << 7 / 9 * 9;
you are doing integer7/9*9 evaluates those numbers as integers, so 7/9 evaluates to 0, and 0*9 = 0.
Equals 0 (the quotient of the division).
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Can a windows message box be display using the cout syntax?
I also need the command prompt window to be suppressed / hidden.
There are ways to call the messagebox function and display text through its usage, but the main constraint here is that cout syntax...
Started by CheeseConQueso on
, 8 posts
by 8 people.
Answer Snippets (Read the full thread at stackoverflow):
Can a windows message....
The c in cout stands * to pass to MessageBox.
No simple way anyway.
You could probably_OK );
See the full docs on MessageBox for more info .
Paradigm, so the cout context isn't really a good one for working with this.
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After performing some tests I noticed that printf is much faster than cout . I know that it's implementation dependent, but on my Linux box printf is 8x faster. So my idea is to mix the two printing methods: I want to use cout for simple prints, and I...
Started by Jabba on
, 12 posts
by 12 people.
Answer Snippets (Read the full thread at stackoverflow):
() { for (int i=0; i<count; i++) { puts(string); puts("\n"); } } void use_cout() { for (int i=0; i<_with_stdio(false); for (int i=0; i<count; i++) std::cout << string << "\n"; std::cout.sync; i++)....
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-edit- I am sending binary and not a string. My test is using html pages so in this example i am only using a string but my question is about binary, vectors and debugging with ostream. I make this clears some confusion.
I have the following code:
cout...
Started by acidzombie24 on
, 6 posts
by 6 people.
Answer Snippets (Read the full thread at stackoverflow):
You can use std::min(n, ostream_iterator<char>(cout) ); total += received; } while( (received > 0) && (total < ); if ( received....
Str(&v[0], n); std::cout << str;
Where n ranges from 0 up to v.size().
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One can remove all calls to printf() using #define printf . What if I have a lot of debug prints like std::cout << x << endl; ? How can I quickly switch off cout << statements in a single file using preprocessor?
Started by Jack on
, 7 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
You may just replace cout by ostream(0) like
#ifdef NDEBUG #define cout ostream(0Substitute your debug output statements with something like this:
IFDBG(cout << result < const&) { return *this; } };
You....
Levels etc..
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