I tried to compile the following code
std::string key = "DISTRIB_DESCRIPTION=";
std::cout << "last five characters: " << key.substr(this.end()-5) << '\n';
And the compiler says
error: invalid use of ‘this’ in non-member function
std::cout << "last five characters: " << key.substr(this.end()-5) << '\n';
^
substr is a "public member function" of std::string, why can't I use this?
I know I could just reference key again instead of this, but my original code was
std::cout << "Description: " << line.substr(found+key.length()+1).substr(this.begin(),this.length()-1) << '\n';
In the second use of substr, the string does not have a name, so the only way to refer to it would be this. I fixed it with
std::cout << "Description: " << line.substr(found+key.length()+1,line.length()-found-key.length()-2) << '\n';
But I am now curious to why this won't work.
this is only available when you are writing code as part of a non-static method of a class. In your particular case, it seems obvious to you that this should refer to key, but the compiler sees no reason for that.
Also, string.substr() takes an integer indicating the beginning position. string.end() returns an iterator, which will not work. What you likely want to do here is call string.length().
Simply replace the first piece of code with:
std::cout << "last five characters: " << key.substr(key.length()-5) << '\n';
And you should be okay.
Related
I'm using Clang to create some internal static code analyzers. For one of the analyzers, we need to take a raw string and check if it has any syntax errors.
We shouldn't consider missing symbols, missing headers, invalid function calls etc. as invalid syntax - as the only meaning is to see if it's a valid C/C++ code or not.
I thought initially that I could do it with ASTUnit:
auto AST = tooling::buildASTFromCodeWithArgs(MyCode,
Args,
"input.cc",
"clang-tool",
std::make_shared<PCHContainerOperations>(),
tooling::getClangStripDependencyFileAdjuster(),
tooling::FileContentMappings(),
&DiagConsumer);
llvm::outs() << "hasUncompilableErrorOccurred " << AST->getDiagnostics().hasUncompilableErrorOccurred() << "\n";
llvm::outs() << "hasUnrecoverableErrorOccurred " << AST->getDiagnostics().hasUnrecoverableErrorOccurred() << "\n";
llvm::outs() << "hasErrorOccurred " << AST->getDiagnostics().hasErrorOccurred() << "\n";
Taking two inputs: Hello world and #include <undefined.h> - both yields 1 in the outputs above - even when #include <undefined.h> is a correct C statement, but the issue with it (unlike with hello world, which's not a valid C code) - is that undefined.h is missing. Similarly, taking: int* p = malloc(sizeof(int)); as code will yield error in all of these calls if stdlib.h wasn't included.
I try to avoid such errors, so that every case, except from hello world, will be considered as valid code.
I did tried to iterate over it by creating a Raw Lexer, but it won't give me sufficient information.
Lexer Lex(CharRange.getBegin(), PP->getLangOpts(), Text.data(),
Text.data(), Text.data() + Text.size());
Token RawTok;
do {
Lex.LexFromRawLexer(RawTok);
llvm::outs() << "\t- " << RawTok.getKind() << "\n";
} while (RawTok.isNot(tok::eof));
I'd love to get any suggestions!
I am currently working on a code and I am trying to use a if-statement on a variable which was taken from a .txt file with a basic string. Its supposed to look like
if (a.variable == "string") {}
When I use
std::cout << a.variable << std::endl;
std::cout << "string" << std::endl;
I get the same results but when using
std::cout << typeid(a.variable).name() << std::endl;
std::cout << typeid("string").name() << std::endl;
I get different results:
NSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEE
and
A5_c.
Could this be the reason why the if-statement failed? Unless I am incorrect, the first typeid stands for a basic string.
I am grateful for any input!
The code I use for reading it looks like:
std::string::size_type beginoption = section.find("=",position);
beginoption = beginoption +1;
std::string::size_type endoption = section.find("\n",position);
optionstorage = section.substr(beginoption, endoption - beginoption);
Two objects in C++ don't have to be the same type to compare as equal. You can compare string objects to string literals because there is an operator== overload that accepts std::string and char const * arguments. (The typeid() operator returns a different value because the two expressions have different types; one is a string object and the other is a char array -- but you can indeed still compare them.)
You mentioned that your "if statement is failing" but when you inspect the contents of the strings, they appear to be the same -- they may actually not be the same. For example, in your code, if a.variable has trailing whitespace, you would not see this in the output and yet the strings would also not be equal.
Try writing both strings surrounded by some characters. I suspect that you will see there is some extra whitespace somewhere:
std::cout << '[' << a.variable << ']' << std::endl;
std::cout << '[' << "string" << ']' << std::endl;
Consider also displaying a.variable.size(). If it's not 6, then the two strings cannot be equal since they have different lengths.
For the purposes of typeid, a.variable is of type std::string while the string literal "string" is of type char const [7].
That explains the output of
std::cout << typeid(a.variable).name() << std::endl;
std::cout << typeid("string").name() << std::endl;
Those code lines:
std::cout << "observerIndex : " <<
std::cout << pobserverIndex -> observerInt() ;
Generate the compiler error below:
file.C:2917:37: error: no match for 'operator<<' (operand types are 'std::basic_ostream<char>' and 'std::basic_ostream<char>')
std::cout << "observerIndex : " <<
^
Could anyone please tell me what left shift operator(<<) is doing on there (before std::cout << pobserverIndex -> observerInt())?
You appear to be missing a semicolon at the end of your first statement, plus you are repeating std::cout.
You need to use
std::cout << "observerIndex: " << pobserverIndex -> observerInt();
A variant like
std::cout << "a" << std::cout<< "b";
is outputting the address of the object cout in the std namespace, formatted as hexadecimal, between the strings "a" and "b".
I am learning C++ and just started reading "Programming Principles and Practice" by Bjarne Stroustrup and he uses this code to illustrate a point:
#include "std_lib_facilities.h"
using namespace std;
int main() // C++ programs start by executing the function main
{
char c = 'x';
int i1 = c;
int i2 = 'x';
char c2 = i1;
cout << c << ' << i1 << ' << c2 << '\n';
return 0;
}
I am familiar in general with the difference between double and single quotes in the C++ world, but would someone kindly explain the construction and purpose of the section ' << i1 << '
Thanks
cout << c << ' << i1 << ' << c2 << '\n';
appears to be a typo in the book. I see it in Programming Principles and Practice Using C++ (Second Edition) Second printing. I do not see it listed in the errata.
According to the book, the intended output is
x 120 x
But what happens here is ' << i1 << ' attempts to compress the << i1 << to a multi-byte character and prints out an integer (most likely 540818464-> 0x203C3C20 -> ASCII values of ' ', '<', '<', ' ') because cout doesn't know wide characters. You'd need wcout for that. End result is output something like
x540818464x
and a warning or two from the compiler because while it's valid C++ code, it's almost certainly not what you want to be doing.
The line should most likely read
cout << c << ' ' << i1 << ' ' << c2 << '\n';
which will output the expected x 120 x
In other words, Linker3000, you are not crazy and not misunderstanding the example code.
Anyone know who I should contact to log errata or get a clarification on the off chance there is some top secret sneakiness going way over my head?
Before answering your question, here is a little background on what that is actually doing. Also note that there is a typo in the example, the string constant should have been double quoted:
cout << c << " << i1 << " << c2 << "\n";
In C++, operators can be overloaded so that they mean different things with different functions. In the case of cout, the << operator is overloaded as the "Insertion Operator". Think of it as taking the operand on the right, and inserting it (or sending it) into the operator on the left.
For example,
cout << "Hello World";
This takes the string "Hello World", and sends it to cout for processing.
So what beginners do not get is what something like this means:
cout << "Hello" << " World";
This is doing the same thing, but the operator precedence says to perform the injections from left to right. To make this work, the cout object returns itself as a function return value. Why is this important? Because the above statement is actually two separate operator evaluations:
(cout << "Hello") << " World";
This first injects "Hello" to cout, which outputs it, then continues to evaluate the next inject operator. Because cout returns itself, after the (cout << "Hello") is executed you have the following still to be evaluated:
cout << " World";
This expression injects " World" into the cout object, which then outputs " World", with the net effect being that you see "Hello World" just like the first time.
So in your example, what is it doing?
cout << c << " << i1 << " << c2 << "\n";
This is evaluated left to right as follows:
((((cout << c) << " << i1 << ") << c2) << "\n"); => Outputs value of c
((((cout ) << " << i1 << ") << c2) << "\n"); => Outputs string " << i1 << "
((( cout ) << c2) << "\n"); => Outputs value of c2
(( cout ) << "\n"); => Outputs newline character
( cout ); => No more output
Expression completes and returns the cout object as the expression value.
Assuming c='x' and c2='x', the final output from this expression is the following character string output on a single line:
x << i1 << x
For beginners, all those insertion operators << look a little strange. It is because you are dealing with objects. You could build the string up as a complete formatted object before injecting it into cout, and while that make the cout expression look simpler, we do not do that in C++ because it makes your code more complex and error prone. Note also, there is nothing special about the cout object. If you wanted to output to the standard error stream, you would use cerr instead. If you wanted to output to a file, your would instantiate a stream object that outputs to the desired file. That rest of the code in your example would be the same.
In C, the same thing would be done procedurally using a format string:
printf("%d << i1 << %d\n", i1, c2);
This is allowed in C++ too, because C++ is a superset of C. Many C++ programmers still use this output method, but that is because those programmers learned C first, and may not have fully embraced the object oriented nature of C++
Note that you may also have seen the << operator in the context of mathematical expressions like:
A = A << 8;
In this case, the << operator is the bitwise rotate operation. It has nothing to do with output to cout. It will rotate the bits in A to the left by eight bits.
I am using Qt, and I have an unsigned char *bytePointer and want to print out a number-value of the current byte. Below is my code, which is meant to give the int-value and the hex-value of the continuous bytes that I receive from a machine attached to the computer:
int byteHex=0;
byteHex = (int)*bytePointer;
qDebug << "\n int: " //this is the main issue here.
<< *bytePointer;
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
}
This gives perfect results, and I get actual numbers, however this code is going into an API and I don't want to use Qt-only functions, such as qDebug. So when I try this:
int byteHex=0;
byteHex = (int)*bytePointer;
std::cout << "\n int: " //I changed qDebug to std::cout
<< *bytePointer;
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
}
The output does give the hex-values perfectly, however the int-values return symbols (like ☺, └, §, to list a few).
My question is: How do I get std::cout to give the same output as qDebug?
EDIT: for some reason the symbols only occur with a certain Qt setting. I have no idea why it happened but it's fixed now.
As others pointed out in comment, you change the outputting to hex, but you do not actually set it back here:
std::cout << " (hex: "
<< std::hex
<< byteHex
<< ")\n";
You will need to apply this afterwards:
std::cout << std::dec;
Standard output streams will output any character type as a character, not a numeric value. To output the numeric value, convert to a non-character integer type:
std::cout << int(*bytePointer);