How to initialize last item automatically in a struct array? - c++

I am passing an array to a function, and i am initializing it globally with some values.
I am using empty string in end of array to determine the array length.
Now, Is there some way to automatically initialize the array to have extra empty item in the end of it, so i have no chances to forget it from there? Just like the char[] works, it adds extra null to the end IIRC.
Here is my code what im using now:
struct twostrings {
string s1, s2;
};
twostrings options[] = {
{"text1", "more text1"},
{"text2", "more text2"},
{"text3", "more text3"},
{""}, // tells that the array ends here
}
int get_len(twostrings opt[]){
int p = 0;
while(1){
if(opt[p].s1 == ""){
return p;
}
p++;
// now here is a possibility to go in infinite loop if i forgot the empty string.
// currently i have a code here that checks if p > 10000 and gives error message to me if i manage to forget that empty string in accident.
}
return p;
}
void dosomething(twostrings options[]){
int len = get_len(options);
for(int p = 0; p < len; p++){
// do stuff
}
}
int main(){ // yes its not valid written main function. dont bother about it.
dosomething(options);
}

Passing around C arrays is not very idiomatic in C++. Try using a std::vector instead:
#include <vector>
#include <string>
struct twostrings {
std::string s1, s2;
};
typedef std::vector<twostrings> option_type;
twostrings options[] = {
{"text1", "more text1"},
{"text2", "more text2"},
{"text3", "more text3"}
};
int get_len(const option_type& options){
return options.size();
}
void dosomething(const option_type& options){
int len = get_len(options);
for(int p = 0; p < len; p++){
// do stuff
}
}
int main() { // This main function is perfectly fine!
option_type opt_vector(options, options + (sizeof options / sizeof options[0]));
dosomething(opt_vector);
}

Unforunately, you're not correct. The char array does not end automatically in a null, this is only a side effect of assigning it with a string literal (which has the automatic null at the end).
char x[] = "ABC"; // size 4, contains A, B, C, \0.
char x[] = {'A','B','C'}; // size 3, contains no terminating null.
So the short answer is no, there's no way to automatically end arrays with an automatic entry. There are a bunch of other options though, such as STL vectors which have other means of determining when you've reached the end. In C++0x there'll probably (IIRC) be a way to initialize the vector just like you'd like.
HTH.
EDIT:
Personally, I prefer to add the extra 0 at the end of the array myself, but I suppose there are ways to work around it using macros.
#define ARRAY(...) {__VA_ARGS__, {0}}
and use it like so
struct foo { char* x; char* y; }
struct foo x[] = ARRAY({"abc", "xyz"}, {"def","uvw"});
I have no idea if this works (and I have no preprocessor handy), and as I said, personally I don't like it. It also requires the first element in the struct to be something which can be assigned 0 to mark the end of the array.
Of course, this forces you to remember to wrap it in the macro call, which is pretty much as bad as forcing you to remember to terminate the array.
EDIT:
I just had a chance to test this and it works. Turns out variadic macros are, so far anyway, C only. However some (most?) C++ compilers support them anyway, a quick search turned up g++ and visual studio. Still I wouldn't favor this approach, I just added it for completeness.

Pass the length or the end instead of using a sentinel:
template<class T, int N>
int len(T (&)[N]) { // exists in a more general form as boost::size
return N;
}
typedef std::pair<std::string, std::string> twostrings;
// std::pairs have first and second members of the given types
void dosomething(twostrings options[], int size);
// call as: dosomething(array, len(array));
# or:
template<class T, int N>
T* end(T (&a)[N]) { // exists in a more general form as boost::end
return a + N;
}
void dosomething(twostrings* options_begin, twooptions* options_end);
// call as: dosomething(array, end(array));
// usage example:
void dosomething(twostrings* options_begin, twooptions* options_end) {
// you might name the parameters just 'begin' and 'end'
for (; options_begin != options_end; ++options_begin) {
// the 'begin' var advances through the entire sequence
// use for (twostrings* current = options_begin; current != options_end; ++current)
// if a separate copy is required
cout << options_begin->first << ": " << options_begin->second << '\n';
}
}
Note the [begin, end) iterator pattern (that's inclusive begin, exclusive end) is common in the stdlib (e.g. look at std::sort from <algorithm>).
This is a good halfway measure between arrays and containers such as std::vector, and allows you to keep the easy initialization syntax you have now (C++0x gives you that same syntax with containers such as std::vector, but 0x is not quite yet ready).

Don't use C style arrays in C++, they're just not worth the effort compared to vector.size(). You should use a boost::array<twostrings, length> for a static array.
Hell, you should probably just not use a static value.

There are better ways of finding array lengths. You can use:
1. sizeof(options) / sizeof(twostrings);
2. sizeof(options) / sizeof(options[0]);
3. std::vector<twostrings> options;
options.size();
4. ARRAYSIZE(options); (windows only)

Btw, if(opt[p].s1 == "") is checking 2 const char * pointers for equality, not 2 strings. Although compiller usualy optimizes equal string constants to point to one place, it is still an error.
You should use a NULL sentinell as it was adviced by Svisstack earlier.
edit: Proof
#include <stdio.h>
const char *one = "the string";
void main(){
const char *other = "the string";
printf("adress of 'one' = %x, it contains \"%s\"\n", one, one);
printf("adress of 'other' = %x, it contains \"%s\"\n", other, other);
if(one == other){
printf("one == other\n", one);
} else {
printf("one != other\n", one);
}
}
Output:
k:\temp>cl test.cpp
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 16.00.30319.01 for 80x86
/out:test.exe
test.obj
k:\temp>test.exe
adress of 'one' = 3d8140, it contains "the string"
adress of 'other' = 3d814c, it contains "the string"
one != other

Related

C++ behavior that I don't understand

My friends and I were playing with the C++ language. While doing so, we encountered something we couldn't understand.
Here is the code:
#include <vector>
#include <iostream>
void print(std::vector<char> const &input)
{
std::cout << input.size();
for (int i = 0; i < input.size(); i++)
{
std::cout << input.at(i) << " - ";
}
}
int main()
{
char cha = 'A';
char chb = 'B';
char * pcha = &cha;
char * pchb = &chb;
try
{
std::vector<char> a = {pcha, pchb};
//std::vector<char> a = {pchb, pcha};
print(a);
}
catch(std::exception e)
{
std::cout << e.what();
}
}
Output for this code:
A
When I comment out this first line try block and uncomment the second line, which comes to this:
try
{
// std::vector<char> a = {pcha, pchb};
std::vector<char> a = {pchb, pcha};
print(a);
}
Output becomes:
std:exception
I thought maybe the this occurs because of the different padding and alignments of the declared variables (char, char*), yet still didn't understand. You can find the code here to play around.
Thanks in advance.
std::vector<char> a = {pcha, pchb};
Here, you use the constructor of vector that accepts two iterators to a range. Unless the end iterator is reachable from the begin one, the behaviour of the program is undefined. Your two pointers are not iterators to the same range (i.e. elements of an array), so one is not reachable from the other. Therefore the behaviour of the program is undefined.
These would be correct:
std::vector<char> a = {cha, chb}; // uses initializer_list constructor
// or
char arr[] {cha, chb};
char * pcha = std::begin(arr);
char * pchb = std::end(arr);
std::vector<char> a = {pcha, pchb}; // uses the iterator constructor
#eerorika's answer explains your mistake.
However, I would like to dissuade you, and other readers, from using the second part of the his(?) corrected code snippet - not because it's incorrect, but because it's problematic coding practice:
I accept Nicolai Jossutis' suggestion of trying to uniformly initialize variables with curly brackets and no equals since (e.g.. mytype myvar {my_initializer};).
Freestanding pointers are dangerous beasts. Try to avoid them altogether, or minimize their existence to where you really need them. After all, you were "tempted" to use those pointers in an inappropriate way... so,
char arr[] {cha, chb};
std::vector<char> a = {std::begin(arr), std::end(arr)};
Don't create a dummy container just to create the one you really want. Just stick with the first line in #eerorika's suggestion (without the equals sign):
std::vector<char> a {cha, chb};
In fact, unless you really need it - you probably don't even want to create a variable-length container. So perhaps just
std::array<char, 2> a {cha, chb};
or with C++17's template argument deduction:
std::array a {cha, chb};

Initializing a Struct's variable const char* const* with C++

I am trying to solve a coding question that requires the results be returned using a given struct. The struct is defined as:
struct Answer
{
const char* const* lastNames;
unsigned numberOfPeople;
}
Where the lastNames is a pointer to last names that are each terminated by a non-alpha char. I can not seem to find any way to convert the vector of strings that I am using to compile all the last names into a variable that I can assign to lastNames. I have tried making a single string with all the last names and assigning it with c_str() like so:
Ans->lastName = allNames.c_str(); but this gives me an error. Due to the limitations of the question I am unable to change the struct variable to anything else. How can I assign a string to a const char* const*
The structure being used effectively uses a C-style approach to defining a variable sized array of pointers to char (with const sprinkled over it). You’ll need storage for both the array of char const* as well as the entities pointed to. Here is how you could build it from a std::vector<std::string>:
std::vector<std::string> strings = somehow_compute_the_strings();
std::vector<char const*> array;
for (std::string const& s: strings) {
array.push_back(s.c_str());
}
Answer answer = { array.data(), array.size() };
Of course, you can’t return answer without the pointer inside pointing to stale data: you’d need to keep the two std::vectors alive. Potentially these two objects could be made members of an object the function is called on. To actually return an object of type Answer without a place to hold on to the std::vectors you could allocate the relevant entities and accept that the result will yield a memory leak unless the caller can clean the result up.
You can't just cast stuff. struct Answer is expecting a char**, so you are going to have to build it and keep it valid as long as the struct Answer is in use. At least they were kind enough to let us know they don't intend to modify it or mess with cleaning up the memory, since it takes "const char * const *".
#include <iostream>
#include <vector>
#include <string>
#include <assert.h>
typedef std::vector<std::string> VectorOfStrings_type;
struct Answer
{
const char* const* lastNames;
unsigned numberOfPeople;
};
class AnswerWrapper
{
private:
// construct and maintain memory so the pointers in the Answer struct will be valid
char ** lastNames;
unsigned int numberOfPeople;
public:
AnswerWrapper(const VectorOfStrings_type &input){
numberOfPeople = input.size();
// create the array of pointers
lastNames = static_cast<char**>(
malloc(numberOfPeople * sizeof(char*))
);
// create each string
for (unsigned int i = 0; i < numberOfPeople; ++i){
const std::string &name = input[i];
// allocate space
lastNames[i] = static_cast<char*>(
malloc(name.size() + 1)
);
// copy string
strncpy(lastNames[i], name.data(), name.size());
// add null terminator
lastNames[i][name.size()] = '\0';
}
}
operator Answer (){
return Answer{ lastNames, numberOfPeople };
}
~AnswerWrapper(){
// critcally important, left as an exercise
assert(0);
}
};
void SomeFunctionWhichUsesAnswer(Answer a){
// presumably you have some legacy C code here
// but here's a quick and easy demo
for (unsigned int i = 0; i < a.numberOfPeople; ++i)
std::cout << a.lastNames[i] << std::endl;
}
int main() {
// Here is your vector of strings
VectorOfStrings_type myData { "custom formatted data goes here", "and more here", "and again" };
// You must construct a buffer for the "Answer" type, which must remain in scope
AnswerWrapper temp{ myData };
// AnswerWrapper is currently in scope, so inside this function, the pointers will be valid
SomeFunctionWhichUsesAnswer(temp);
}
Also, I noticed that the strings in Answer are not referred to as null terminated. That is a separate issue you can take care of.
A const member variable can only be assigned in the constructor.
if you can add to the struct, define a constructor, and use the : lastname(value) syntax; or use the struct Answer myVar{value,number}; initialization, right where you declare your instance.
Another - ugly, dangerous, and frowned upon - alternative is a cast: (char**) lastname = value;, or in C++ syntax reinterpret_cast<char**>(lastname) = value.
If someone is teaching you either of those approaches, change the teacher.

function that returns an array of objects

I have got a structure
class pyatno {
int pyatnoNumber;
int locX, locY;
bool possible;
char *number;
char pyatnoView[4][4];
}
the idea is to make a function, that would return an array of pyatno.pyatnoView objects, but there is a mess. I don't understand how exactly I can get access to this "property". I am not strong in c++, so if it isn't real, or i am talking something wrong, explain please, cause I am really stacked in this question.
As you mentioned that you are not very strong with c++, and your question is rather unclear, here are several suggestions.
To get access to a class's attributes, c++ has the notion of visibility; The default visibility is private, that is, attributes and functions will not be visible outside of the class:
class Foo {
int some_value;
};
There are several ways you can retrieve data from an object, however to put it simply, you should either make the attribute public:
class Foo {
public:
int some_value;
};
or expose it via accessors/mutators:
class Foo {
int some_value;
public:
int get_some_value() { return some_value; }
void set_some_value(int v) { some_value = v; }
};
Another thing to note is that you can not return arrays! In c++, when an array passes a function boundary (that is to say, passed as a parameter to, or returned from), and in a lot of other cases, an array will 'decay' in to a pointer. For example, the following is how I would pass an array of characters (otherwise known as a c-string) to a function:
#include <iostream>
using namespace std;
void print_cstr(const char *cstr) {
cout << cstr << endl;
}
int main() {
const char my_cstr[20] = "foo bar baz qux";
print_cstr(my_cstr);
return 0;
}
So what happens for N-dimensional arrays? Well, if char[1] decays to char*, then char[1][1] will decay to char**, and so on. You might have noticed this with the older main signature in C programs, which is used to pass an array of strings representing arguments passed to the program:
int main(int argc, char **argv) { ... }
It is very important that you realise that this really is no longer an array. C style strings are a bit special, in that they are conventionally terminated with a null byte \0, which means that you can usually tell where the end of the string is, or how long it is. However, you no longer have any information on how long the array is! For example, this is completely legal:
#include <iostream>
using namespace std;
void bad_fn(const int *nums) {
for (unsigned i = 0; i < 20; ++i) {
cout << "num " << i << " = " << nums[i] << endl;
}
}
int main() {
const int my_nums[5] = { 1, 2, 3, 4, 5, };
bad_fn(my_nums);
return 0;
}
Your function will end up reading memory beyond the bounds of the array, as it has no way of knowing where the array begins or ends (after all, array indexes are just pointer arithmetic). If you do not want to have to worry about keeping track of, and passing around the length of your array (and I would suggest that you do not!), please look at using one of the C++ standard library's containers. std::vector and std::array are two examples that would fit in the use case you have provided, and you can find decent documentation for them here.

How to make char array and std::string "in a relationship"?

I'm looking for a way to associate a char array with a string so that whenever the char array changes, the string also changes. I tried to put both char array and string variables in a union but that didn't worked as the compiler complained...
Any ideas are welcome...
class Observable_CharArray
{
char* arr;
std::function<void(char*)> change_callback;
public:
Observable_CharArray(int size, std::function<void(char*)> callback)
: arr(new char[size]), change_callback(callback){}
~Observable_CharArray()/*as mentioned by Hulk*/
{
delete[] arr;
}
void SetCallback(std::function<void(char*)> callback)
{
change_callback = callback;
}
/*other member function to give access to array*/
void change_function()
{
//change the array here
change_callback(arr);
}
};
class Observer_String
{
std::string rep;
void callback(char* cc)
{
rep = std::string(cc);
}
public:
Observer_String(Observable_CharArray* och)
{
och->SetCallback(std::bind(&callback, this, _1));
}
/*other member functions to access rep*/
};
The design can definitely be improved.
There can be other ways to solve your actual problem rather than observing char arrays.
The problem is that the std::string may change the string array inside (especially when it resizes). For instance, c_str returns the address of the current string - documentation says that "The pointer returned may be invalidated by further calls to other member functions that modify the object.".
If you're sure you won't call string methods (hence the string will stay at the same memory location), you could try accessing the c_str pointer (your char array) directly and modify its content.
std::string str = "test";
char* arr = (char*)str.c_str();
arr[3] = 'a';
NOTE: I strongly advice against this unless in a testing context.
In other words, the string class doesn't guarantee it's going to stay in the same place in memory - meaning trying to access it through a char array is impossible.
The best is to create another string class that enforces the char array to always stay the same size (and so can stay in the same memory position all the time). You could also create a bigger array (max size string for instance) to cope with any string size changes - but that should be enforced in your wrapper class.
Well you can do this, but you shouldn't
#include <iostream>
#include <string>
int main()
{
std::string test("123456789");
std::cout << test << "\n";
char* data = &test.front(); // use &(*test.begin()) for pre-C++11 code
for ( size_t i(0); i < test.size(); ++i )
{
data[i] = 57 - i;
}
std::cout << test << "\n";
}
Output will be
123456789
987654321
This however goes again everything std::string is trying to facilitate for you. If you use data, you risk causing UB and changes to test may make data point to garbage.
You should not do this!
However, there are many (dangerous) ways to achieve it:
char* cStr = const_cast<char*>(cppStr.c_str());
or
char* cStr = const_cast<char*>(cppStr.data());
or
char* cStr = &cppStr[0];
But beware that the cppStr might be reallocated whenever you touch it, hence invalidating your cStr. That would crash at some point in time, although maybe not immediately (which is even worse).
Therefore, if you are going to do this anyway. Make sure to cppStr.reserve(SOMETHING) *before* you get the cStr out of it. This way, you will at least stabilise the pointer for a while.

What's the best way to rewrite this generic function I wrote in C++ in C?

//Prints out a given array
template <typename T>
void print(T t)
{
for(int i = 0; i < t.size(); i++)
{
cout << t[i] << " ";
}
cout << endl;
}
I have an idea but it includes passing the size of the array. Is it possible to avoid this?
*Update
Thanks for all of the answers/ideas but this problem is getting way deeper than my snorkeler can handle. I wanted to rewrite my C++ code in C because it was horribly written and slow. I see now that I have an opportunity to make it even worse in C. I'll rewrite it from the ground up in Python(performance be damned). Thanks again
If you don't have ELEMENTS, it's
#define ELEMENTS(a) (sizeof(a)/sizeof(*a))
Then,
#define print_array(a, specifier) print_array_impl(a, specifier, ELEMENTS(a), sizeof(*a))
void print_array_impl(void* a, char* specifier, size_t asize, size_t elsize)
{
for(int i = 0; i < asize; i++)
{
// corrected based on comment -- unfortunately, not as general
if (strcmp(specifier, "%d") == 0)
printf(specifier, ((int*)a)[i]);
// else if ... // check other specifiers
printf(" ");
}
printf("\n");
}
Use like this
print_array(a, "%d") // if a is a int[]
and, a needs to be an array name, not a pointer (or else ELEMENTS won't work)
You cannot know what is the size of an array without passing the size of that array (except operating with sizeof in static arrays). This is because the a pointer to a block of memory will only point to the base of the block of memory, from which you can know where the array/block of memory starts, but as there is no end defined you cannot determine where it will end.
You either need to set your own length per array and preserve it, and use it with the array like as described:
You can make a new type like:
struct _my_array {
typename arr[MAX];
int n;
} my_array;
OR
struct _my_array {
typename *arr;
int n;
} my_array;
In this case you need to allocate the a block of memory dynamically with new or malloc , and when finished free the memory with delete or free (respectively).
Or you can simply pass the array number of elements through the function.
Another way is to use a special terminator value of your array type which if encountered will be determined as the end of the array. In this case you need not preserve the size. For example a string is '\0' terminated, so all the string functions know that when a '\0' character is encounter in the char array it will consider that the string has end.
UPDATE
Because this is a generic function and the array can be of any type, one thing which you can do is like this:
struct _my_generic_arr {
void *arr;
int n;
int type;
} my_generic_arr;
When populating this array you can use any type. To identify which type, pass an identified in the type component. Each unique value will determine which type does the arr pointer actually points to (was actually the intended type to be pointed). The n will define the length. Now, depending on different values of type make a switch - case or an if - else ladder or nest, and process the array as you need.
It is impossible in c to track the size of an array in other block,,
It would be a better option to pass the size of the array along..
The other option would be to declare a global variable that has the size and using that variable inside the function
Eg,,
int size=<some value>
void main()
{
int arr[<same value>];
}
void print(T t)
{
for(int i = 0; i < size; i++)
{
printf("%d ",t[i]) //assuming T as int
}
printf("\n");
}
In C, you would need to pass two additional parameters: the size of the array (as you mentioned), and some way of indicating how to convert t[i] into a string. To convert t[i] to a string, you could create a custom switch statement to decode possible types, pass a pointer to a function that will return the string pointer, or you could pass the printf format specifier (e.g. "%d" for integer).
The problem is larger than you think. If you have an array of size 12, how do you know what data is held in that array? It could be 3 char*'s (on 32 bit system), 3 int32_t's, or even 12 chars. You have no way of knowing how to interpret the data. The best you could do is to implement your own version of a v-table and putting a print or toString function into it.
typedef struct {
void *array;
size_t length;
int element_width;
printer_t to_string;
} container;
printer_t is a type that describes a function pointer that takes an element pointer and returns a string (or prints it, if you don't want to free the string). This is almost never worth doing in C. That doesn't mean it can't be done. I would emphasize, though, that none of this is intended to imply that it should be done.
The function itself would look something like this:
void print(container *thing)
{
size_t offset;
int width;
char *stringified;
width = thing->element_width;
for (offset = 0; offset * width < thing->length; offset += width)
{
stringified = thing->to_string(thing->array + offset);
printf("%s ", stringified);
free(stringified);
}
}
What this does is essentially turn a struct into a faux class with a function pointer for a method. You could be more object-oriented and put the method in the type being printed and make it an array of those instead. Either way, it's not a good idea. C is for writing C. If you try to write in a different language, you'll end up with all sorts of terrible stuff like this.