`sizeof(bool)` is implementation defined, and the standard puts notable emphasis on this fact.

&#167;5.3.3/1, abridged:

&gt; `sizeof(char)`, `sizeof(signed char)` and `sizeof(unsigned char)` are 1; the result of `sizeof` applied to any other fundamental type is implementation-defined. [Note: in particular, `sizeof(bool)` and `sizeof(wchar_t)` are implementation-defined.&lt;sup&gt;69)&lt;/sup&gt;]

Footnote 69):

&gt; `sizeof(bool)` is not required to be 1.

http://msdn.microsoft.com/en-us/library/tf4dy80a.aspx

&quot;In Visual C++4.2, the Standard C++ header files contained a typedef that equated bool with int. In Visual C++ 5.0 and later, bool is implemented as a built-in type with a size of 1 byte. That means that for Visual C++ 4.2, a call of sizeof(bool) yields 4, while in Visual C++ 5.0 and later, the same call yields 1. This can cause memory corruption problems if you have defined structure members of type bool in Visual C++ 4.2 and are mixing object files (OBJ) and/or DLLs built with the 4.2 and 5.0 or later compilers.&quot;

It&#39;s implementation defined. Only `sizeof(char)` is `1` by the standard.

See 5.3.3 paragraph 1 :

&gt; [Note: in particular, sizeof(bool) and
&gt; sizeof(wchar_t) are
&gt; implementation-defined.69) ]



I&#39;m using an open source library which seems to have lots of preprocessing directives to support many languages other than C. So that I can study what the library is doing I&#39;d like to see the C code that I&#39;m compiling after preprocessing, more like what I&#39;d write.

Can gcc (or any other tool commonly available in Linux) read this library but output C code that has the preprocessing converted to whatever and is also readable by a human?

Can gcc output C code after preprocessing?

I am writing an FTP downloader. Part of to the code is something like this:

    ftp.retrbinary(&quot;RETR &quot; + file_name, process)

I am calling function process to handle the callback:

    def process(data):
	    print os.path.getsize(file_name)/1024, &#39;KB / &#39;, size, &#39;KB downloaded!&#39;
	    file.write(data)

and output is something like this:

    1784  KB / KB 1829 downloaded!
    1788  KB / KB 1829 downloaded!
    etc...   

but I want it to print this line and next time reprint/refresh it so it will only show it once and I will see progress of that download.

How can it be done?

Output to the same line overwriting previous output?

I can&#39;t find an answer in the standard documentation. Does the C++ language standard require `sizeof(bool)` to always be 1 (for 1 byte), or is this size implementation-defined?

Is sizeof(bool) defined in the C++ language standard?

<p>I can't find an answer in the standard documentation. Does the C++ language standard require <code>sizeof(bool)</code> to always be 1 (for 1 byte), or is this size implementation-defined?</p>


I am trying to write into a file and if the file doesn&#39;t exist create it. I have searched on the internet and nothing worked for me.

My code looks currently like this:

    QString filename=&quot;Data.txt&quot;;
    QFile file( filename );
    if ( file.open(QIODevice::ReadWrite) )
    {
        QTextStream stream( &amp;file );
        stream &lt;&lt; &quot;something&quot; &lt;&lt; endl;
    }

If I create a text file called Data in the directory, it remains empty. If I don&#39;t create anything it doesn&#39;t create the file either.
I don&#39;t know what to do with this, this isn&#39;t the first way in which I tried to create/write into a file and none of the ways worked.

Thanks for your answers.

Creating/writing into a new file in Qt

As we know if `n` is not a perfect square, then `sqrt(n)` would not be an integer. Since I need only the integer part, I feel that calling `sqrt(n)` wouldn&#39;t be that fast, as it takes time to calculate the fractional part also. 

So my question is, 

Can we get only the integer part of *sqrt(n)* without calculating the actual value of `sqrt(n)`? The algorithm should be faster than `sqrt(n)` (defined in `&lt;math.h&gt;` or `&lt;cmath&gt;`)?

If possible, you can write the code in `asm` block also.

Fastest way to get the integer part of sqrt(n)?

I have a template function with one argument.
I have to instantiate that function without calling that function means explicitly I have to instantiate.

I have this function:

    template &lt;class T&gt; int function_name(T a) {}

I instantiated that function like this:

    template int function_name&lt;int&gt;(int);

But I got the following errors:

    error: expected primary-expression before &#39;template&#39;
    error: expected `;&#39; before &#39;template&#39;


How do I explicitly instantiate a template function?

A Base Class pointer can point to a derived class object. Why is the vice-versa not true without casting?
Logically a base class would not have enough information of the derived class but a derived class should have the information of the base class as well.
I am missing some basics here.

A Base Class pointer can point to a derived class object. Why is the vice-versa not true?

I guess not, but I would like to confirm. Is there any use for `const Foo&amp;&amp;`, where `Foo` is a class type?

Do rvalue references to const have any use?

The following code doesn&#39;t print out anything:

    $bool_val = (bool)false;
    echo $bool_val;

But the following code prints `1`:

    $bool_val = (bool)true;
    echo $bool_val;

Is there a better way to print `0` or `false` when `$bool_val` is `false` than adding an `if` statement?

PHP - Get bool to echo false when false

I have a jQuery function tied to my submit button like this:

    $(function () {
        $(&#39;#signupform&#39;).submit(function () {
            alert(&#39;test&#39;);
        });
    });

However, it fires whether or not the form is valid. My model is decorated with various DataAnnotations and the client-side validation is working well, but I only want that jQuery function to fire if the form has been validated. How do I accomplish that?

**EDIT:** To clarify, I&#39;m using MVC DataAnnotations + jQuery&#39;s unobtrusive javascript to handle the client-side validation. I do not have my own javascript validation routines written. The built in jQuery validation is doing a great job of validating the form, but I just need to know how to get the results of that validation into a boolean value within my own function.

How to fire jQuery function only if form is valid

I tried running this piece of code:

    path = &#39;/bla/bla/bla&#39;

    if path is True:
        print &quot;True&quot;
    else:
        print &quot;False&quot;

And it prints **False**. I thought Python treats anything with value as **True**. Why is this happening?


&#39;True&#39; and &#39;False&#39; in Python

I was wondering, does `true` equal to 1 and `false` equal to 0 and how?

Does true equal to 1 and false equal to 0?

I want to write a bash function that check if a file has certain properties and returns true or false. Then I can use it in my scripts in the &quot;if&quot;. But what should I return?

    function myfun(){ ... return 0; else return 1; fi;}

then I use it like this:

    if myfun filename.txt; then ...

of course this doesn&#39;t work. How can this be accomplished?

Returning a boolean from a Bash function

The below fails to compile:

    typedef int arr[10];
    int main(void) {
    	return sizeof arr;
    }

    sizeof.c:3: error: expected expression before ‘arr’
but if I change it to

    sizeof(arr);
everything is fine. Why?

Why (and when) do I need to use parentheses after sizeof?

Is the size of a pointer the same as the size as the type it&#39;s pointing to, or do pointers always have a fixed size?  For example...

    int x = 10;
    int * xPtr = &amp;x;
    char y = &#39;a&#39;;
    char * yPtr = &amp;y;

    std::cout &lt;&lt; sizeof(x) &lt;&lt; &quot;\n&quot;;
    std::cout &lt;&lt; sizeof(xPtr) &lt;&lt; &quot;\n&quot;;
    std::cout &lt;&lt; sizeof(y) &lt;&lt; &quot;\n&quot;;
    std::cout &lt;&lt; sizeof(yPtr) &lt;&lt; &quot;\n&quot;;

What would the output of this be?  Would `sizeof(xPtr)` return 4 and `sizeof(yPtr)` return 1, or would the 2 pointers actually return the same size?

The reason I ask this is because the pointers are storing a memory address and not the values of their respective stored addresses.





What is the size of a pointer?

This is a C++ interview test question not homework. 

    #include &lt;iostream&gt;
    using namespace std;
    enum months_t { january, february, march, april, may, june, july, august, september,    
      october, november, december} y2k;

     int main ()
      {
        cout &lt;&lt; &quot;sizeof months_t is  &quot; &lt;&lt; sizeof(months_t) &lt;&lt; endl;
        cout &lt;&lt; &quot;sizeof y2k is  &quot; &lt;&lt; sizeof(y2k) &lt;&lt; endl;
        enum months_t1 { january, february, march, april, may, june, july, august,    
           september, october, november, december} y2k1;
        cout &lt;&lt; &quot;sizeof months_t1 is  &quot; &lt;&lt; sizeof(months_t1) &lt;&lt; endl;
        cout &lt;&lt; &quot;sizeof y2k1 is  &quot; &lt;&lt; sizeof(y2k1) &lt;&lt; endl;
     }

Output:
&gt; sizeof months_t is  4  
&gt; sizeof y2k is  4  
&gt; sizeof months_t1 is  4  
&gt; sizeof y2k1 is  4  

Why is the size of all of these 4 bytes? Not 12 x 4 = 48 bytes?  
I know union elements occupy the same memory location, but this is an enum.


what is the size of an enum type data in C++?

Here is the code compiled in dev c++ windows:

    #include &lt;stdio.h&gt;
    
    int main() {
        int x = 5;
        printf(&quot;%d and &quot;, sizeof(x++)); // note 1
        printf(&quot;%d\n&quot;, x); // note 2
        return 0;
    }

I expect `x` to be 6 after executing **note 1**. However, the output is:

    4 and 5

Can anyone explain why `x` does not increment after **note 1**?

Why does sizeof(x++) not increment x?

I&#39;m trying to use 64 bit integers in C, but am getting mixed signals as to whether it should be possible.

When I execute the printf:

    printf(&quot;Size of long int:%d\nSize of long long int:%d\n\n&quot;,(int)sizeof(long int), (int)sizeof(long long int));

The response I get is:

Size of long int:4
Size of long long int:8

This makes me feel that a long long int has 8 bytes = 64 bits.

However, when I try to declare the following variables:

    long long int a2 = 0x00004444;
    long long int b2 = 0x000044440;
    long long int c2 = 0x0000444400;
    long long int d2 = 0x00004444000;
    long long int e2 = 0x000044440000;
    long long int f2 = 0x0000444400004;
    long long int g2 = 0x00004444000044;
    long long int h2 = 0x000044440000444;
    long long int i2 = 0x0000444400004444;

The last 4 variables (f2,g2,h2,i2) give me the error message:  

warning: integer constant is too large for ‘long’ type 

I get the same result when I replace &#39;long long int&#39; with &#39;int64_t&#39;. I assume &#39;int64_t&#39; was recognized, since it didn&#39;t generate any error messages of its own.

So, it appears my 8 byte long long int is really a 6 byte long long int, and I don&#39;t understand what I&#39;m missing here. If it&#39;s any help, here is the information on my gcc compiler:
 

    me@ubuntu:~$ gcc -v  
    Using built-in specs.  
    Target: i686-linux-gnu  
    Configured with: ../src/configure -v   
    --with-pkgversion=&#39;Ubuntu/Linaro 4.4.4-14ubuntu5&#39;  
    --with-bugurl=file:///usr/share/doc/gcc-4.4/README.Bugs   
    --enable-languages=c,c++,fortran,objc,obj-c++  
    --prefix=/usr   
    --program-suffix=-4.4   
    --enable-shared   
    --enable-multiarch   
    --enable-linker-build-id   
    --with-system-zlib   
    --libexecdir=/usr/lib   
    --without-included-gettext   
    --enable-threads=posix   
    --with-gxx-include-dir=/usr/include/c++/4.4   
    --libdir=/usr/lib   
    --enable-nls   
    --with-sysroot=/ -  
    -enable-clocale=gnu   
    --enable-libstdcxx-debug   
    --enable-objc-gc   
    --enable-targets=all 
    --disable-werror   
    --with-arch-32=i686   
    --with-tune=generic   
    --enable-checking=release   
    --build=i686-linux-gnu   
    --host=i686-linux-gnu   
    --target=i686-linux-gnu  
    Thread model: posix  
    gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5)   

If anyone knows how (or if) 64 bit integers are accessible to me, I&#39;d really appreciate any help. Thanks....







How to specify 64 bit integers in c

I posted a question with my code whose only `#include` directive was the following:

    #include &lt;bits/stdc++.h&gt;

My teacher told me to do this, but in the comments section I was informed that I shouldn&#39;t.

Why?

Why should I not #include &lt;bits/stdc++.h&gt;?

I saw the below code in [this Quora post](https://www.quora.com/What-is-something-that-almost-nobody-knows-about-the-C-coding-language/answer/%C3%81lvaro-Lopes):

    #include &lt;stdio.h&gt;
     
    struct mystruct { int enabled:1; };
    int main()
    {
      struct mystruct s;
      s.enabled = 1;
      if(s.enabled == 1)
        printf(&quot;Is enabled\n&quot;); // --&gt; we think this to be printed
      else
        printf(&quot;Is disabled !!\n&quot;);
    }

In both C &amp; C++, the output of the code is *unexpected*, 

&gt; Is disabled  !!

Though the &quot;sign bit&quot; related explanation is given in that post, I am unable to understand, how it is possible that we set something and then it doesn&#39;t reflect as it is.

Can someone give a more elaborate explanation?

---

**Note**: Both the tags [tag:c] &amp; [tag:c++] are required, because their standards slightly differ for describing the bit-fields. See answers for [C specification](https://stackoverflow.com/a/53853737/514235) and [C++ specification](https://stackoverflow.com/a/53867011/514235).

Content Type	Original Author	Original Content on Stackoverflow
Question	0xbadf00d	View Question on Stackoverflow
Solution 1 - C++	GManNickG	View Answer on Stackoverflow
Solution 2 - C++	Yi Ling	View Answer on Stackoverflow
Solution 3 - C++	peoro	View Answer on Stackoverflow
Solution 4 - C++	BЈовић	View Answer on Stackoverflow

Is sizeof(bool) defined in the C++ language standard?

C++ Problem Overview

C++ Solutions

Solution 1 - C++

Solution 2 - C++

Solution 3 - C++

Solution 4 - C++

Output to the same line overwriting previous output?

Can gcc output C code after preprocessing?

Attributions