I want to parse a json file/string with boosts property_tree, but instead of having sub-trees parsed into an array I would like it to stay as a string for use in another existing function which only deals with json-Strings.
I hope the following example is sufficient:
example.json
{
"type": "myType",
"colors": {
"color0":"red",
"color1":"green",
"color2":"blue"
}
}
main.cpp
std::stringstream ss("example.json");
ptree pt;
read_json(ss, pt);
std::string sType = pt.get("type", "");
std::string sColors = pt.get<std::string>("colors");
std::cout << "sType: " << sType << std::endl; // sType: myType
std::cout << "sColors: " << sColors << std::endl; // sColors: {"color0":"red", "color1":"green", "color2":"blue"}
I've tried several functions, for example pt.get_child("colors") would just return another ptree and pt.get_value<std::string>("colors") does return an empty string ("").
The desired output would look like this:
sColors: {"color0":"red", "color1":"green", "color2":"blue"}
or
sColors: {\"color0\":\"red\", \"color1\":\"green\", \"color2\":\"blue\"}
Is there a way to recive the desired output for sColors?
I found a possible solution faster than anticipated, the following code will provide a satisfactory answer:
std::stringstream os;
write_json(os, pt.get_child("colors"), false);
std::string sColors = os.str();
std::cout << "sColors: " << sColors << std::endl;
If there is a more elegant solution feel free to post it as well!
Related
File:
{
"somestring":{
"a":1,
"b":7,
"c":17,
"d":137,
"e":"Republic"
},
}
how can I read the somestring value by jsoncpp?
Use the getMemberNames() method.
Json::Value root;
root << jsonString;
Json::Value::Members propNames = root.getMemberNames();
std::string firstProp = propNames[0];
std::cout << firstProp << '\n'; // should print somestring
If you want to see all the properties, you can loop through it using an iterator:
for (auto it: propNames) {
cout << "Property: " << *it << " Value: " << root[*it].asString() << "\n";
}
This simple loop will only work for properties whose values are strings. If you want to handle nested objects, like in your example, you'll need to make it recursive, which I'm leaving as an exercise for the reader.
This question already has answers here:
Why fill_n() does not work with vector.reserve()?
(3 answers)
Closed 5 years ago.
I'm trying to transform a vector of foo instances to a string but I'm having fatal error on std::transform.
Say data has the following value:
[0]
[name] = John
[size] = 3423
[1]
[name] = Joseph
[size] = 3413
Code:
struct foo {
foo(std::string n, size_t s)
: name(std::move(n)),
size(s)
{
}
std::string name;
size_t size;
};
std::string server = "1";
std::vector<std::string> output;
output.reserve(static_cast<unsigned_long>(std::distance(std::begin(data), std::end(data))));
std::transform(std::begin(data),
std::end(data),
std::begin(output),
[&, this](foo const& item){
std::ostringstream result;
data << server << ","
<< item.name << ","
<< item.size << ";";
return result.str();
});
While debugging it stops at the line
*__result = __unary_op(*_first) of tranform implementation in stl_algo.h then goes to FatalConditionHandler of catch test framework. I'm new to both catch testing and std::transform. Can someone explain what might cause the problem and how to solve it? Thanks a lot!
You've reserved space in output, but you've left its size at zero.
You then proceed to write through its begin iterator, just as if it had space to hold data.
Then everything goes "boom".
Instead of writing through std:begin(output), consider using std::back_inserter(output) as the destination iterator.
You also have one other problem: inside your lambda, you have:
std::ostringstream result;
data << server << ","
<< item.name << ","
<< item.size << ";";
return result.str();
This looks like a fairly obvious mistake--you undoubtedly intended:
std::ostringstream result;
result << server << ","
<< item.name << ","
<< item.size << ";";
return result.str();
Personally, I'd probably structure the code somewhat differently. I'd add something like:
struct foo {
std::string name;
size_t size;
// new addition:
friend std::ostream &operator<<(std::ostream &os, foo const &f) {
return os << f.name << ',' << f.size;
}
};
...then the lambda in your transform becomes rather simpler:
std::ostringstream result;
result << server << "," item;
return result.str();
It might, however, be worth considering doing this without the stringstream itermediaries at all. In this case, you really just need string concatenation, and they impose quite a bit of overhead to do that.
struct foo {
// ...
std::string to_string() {
return name + "," + size;
}
};
Then the lambda body becomes:
return server + "," + item.to_string();
Shorter, simpler, and almost certainly faster.
I need a C++ function that dumps some text data (multiple lines) to the console:
void DumpData() const
{
std::cout << "My Data 1" << std::endl;
std::cout << "My Data 2" << std::endl;
}
This should be the default behaviour, however, it must also be possible to pass some other stream object that would be used instead of std::cout, something like this:
void DumpData(std::ostream& Stream = std::cout) const
{
Stream << "My Data 1" << std::endl;
Stream << "My Data 2" << std::endl;
}
Now to my questions:
What is the correct type I should use for the paramter (std::ostream& in this example)?
What is the default value, can I use = std::cout directly?
Moreover (3.), after the call to this function, if I pass my own stream object, I need to iterate over all strings in this stream line by line. What would be the best way to achieve this?
Why don't you just try it yourself?
Here's you code in Coliru for std::cout and std::stringstream as an example (constness of DumpData removed obviously):
#include <iostream>
#include <sstream>
void DumpData(std::ostream& Stream = std::cout)
{
Stream << "My Data 1" << std::endl;
Stream << "My Data 2" << std::endl;
}
int main() {
DumpData();
std::stringstream ss;
DumpData(ss);
std::string l;
while(std::getline(ss, l)) {
std::cout << l << std::endl;
}
return 0;
}
Output is what you expected.
1 and 2 are correct. Your other option is to use std::ostringstream, but since std::cout is a std::ostream you would need to define another function with this signature.
To iterate the custom output, I would convert the stream to a string, then use some kind of string splitting to read each line.
Adding to #Jay 's answer, you could use a template parameter to be able to use a variety of streams such as std::stringstream or a std::iostream as long as the template parameter supports the << operator.
template <typename T> // = decltype(something)
void DumpData(T& Stream = std::cout) const
{
Stream << "My Data 1" << std::endl;
Stream << "My Data 2" << std::endl;
}
You can take it one step further by ensuring the type T provided overloads the operator <<.
Also in some cases, certain (possibly custom) streams may not be able to support std::endl so it might be safer to default to using \n which is always nicer since it avoid unnecessary flushes.
I often end up writing code like this:
SomeStreamableType x;
std::stringstream ss;
ss << "Value is: " << x;
log(ss.str());
The extra line needed to generate the stringstream feels superfulous. I can do this, but it's equally cumbersom:
SomeStreamableType x;
const std::string str = "Value is: " + boost::lexical_cast<std::string>(x);
log(str);
I want to be able to do this:
SomeStreamableType x;
log(std::stringstream() << "Value is: " << x);
Have others encountered this issue and come up with a workaround? I don't want to create any helper functions or classes.
Your code will work without modifications, as long as log accepts an ostream& reference:
void log(ostream& o) {
stringstream* s = dynamic_cast<stringstream*>(&o);
if (s) {
cout << s->str() << endl;
}
}
int main() {
int x = 5, y = 6;
log(stringstream() << "x=" << x << ", y=" << y);
return 0;
}
Demo.
To solve this problem I have often simply done something like this:
#define LOG(m) do{std::ostringstream oss;oss<<m;std::cout<<oss.str()<<'\n';}while(0)
// ...
LOG("some text: " << value1 << ' ' << value2); // no need for '\n'
Now I tend to use a more sophisticated class based solution that has an even nicer interface and doesn't use a horrible macro.
Hi below is my function:
string Employee::get_print(void) {
string out_string;
stringstream ss;
ss << e_id << " " << type << endl;
out_string = ss.str();
return out_string;
}
e_id and type are int and they contain values from the class Employee. But when I pass them into the stringstream they just clear the string when I try to out put it. But if I don't have a int in the ss << "Some text" << endl; this output fine. What am I doing wrong =S
//Edit
Ok;
This is the calling code:
tmp = cur->get_print();
Where tmp is a string and cur is an Employee Object.
This code...
stringstream out;
out << "Test " << e_id << " " << e_type;
return out.str();
Retruns "Test " and nothing else. If I take out "Test " << my returned string is ""
I'm using GCC 4.2 on Mac OS/X 10.6.2 if that makes any difference.
I too am unable to reproduce this error. As has been mentioned, don't include the endl, as this actually appends a \n and is supposed to flush the write buffer. For this use, it is completely unnecessary and may actually lead to undesirable results...However, the code in your edit/update works just fine for me.
int main(int argc, char* argv[])
{
int e_id = 5;
int e_type = 123456;
stringstream out;
out << "Test " << e_id << " " << e_type;
cout << out.str();
return 0;
}
Produces:
Test 5 123456
My suggestions would be to double check that e_id and e_type are really just native int.
For further testing, you may want to force a cast on the values to see if it helps as such:
out << "Test " << (int)e_id << " " << (int)e_type;
Since I'm unable to reproduce this error, I'm afraid I'm unable to help any further. But best of luck to you!
Ok I have no idea what is going on with stringstream I've tried using it in other parts of my code and it doesn't work with integers. Therefore, I have reverted to using the sprintf C function:
string Employee::get_print(void) {
char out[50];
sprintf(out, "%d %d", e_id, e_type);
string output = out;
return output;
}
This returns the string which is needed.
I have moved into Netbeans and I don't have this problem. So it is an issue with Xcode.
I think the endl is not needed. You only need to write endl if you want to write a newline on a file on on std::cout.
Since you write endl, your stringstream will contain a string with 2 lines of which the second is empty. This probably confuses you. Remove the endl to get only one line.
I've got exactly the same problem - GCC and stringstream returning nothing.
As I found out, the trick is that you have to put some text data before anything else into the stringstream.
This code...
stringstream ss(stringstream::out);
ss << 3.14159;
cout << "'" << ss.str() << "'" << endl;
gets you this result:
''
But if we put a single "" inside the stringstream first...
stringstream ss(stringstream::out);
ss << "" << 3.14159;
cout << "'" << ss.str() << "'" << endl;
the result is as expected:
'3.14159'