C++ Printing array of struct objects
0
I have a problem with printing an array of struct objects.
Here is the code:
#include <iostream>
#include <string>
#define LOG(x) std::cout << x
#define ARR_SIZE(arr) sizeof(arr)/sizeof(arr[0])
struct vec2
{
int x, y;
};
struct Painting : vec2
{
std::string s_author, s_title, s_genre;
};
void disp_Array(Painting arr)
{
for (int i = 0; i < ARR_SIZE(arr); i++)
{
LOG(arr[i].s_author << "n");
LOG(arr[i].s_title << "n");
LOG(arr[i].s_genre << "n");
LOG(arr[i].x << "n");
LOG(arr[i].y << "n");
}
}
int main()
{
Painting painting[4];
disp_Array(painting);
for (int i = 0, j = 1; i < ARR_SIZE(painting), j <= ARR_SIZE(painting); i++, j++)
{
LOG("Painting " << j << "n");
LOG("Author: ");
std::cin.ignore();
std::getline(std::cin, painting[i].s_author);
LOG("Title: ");
std::getline(std::cin, painting[i].s_title);
LOG("Genre: ");
std::getline(std::cin, painting[i].s_genre);
LOG("Width: ");
std::cin >> painting[i].x;
LOG("Height: ");
std::cin >> painting[i].y;
LOG("n");
}
disp_Array(painting);
std::cin.get();
return 0;
}
The problem is within the disp_Array function. The idea was that the function should output the variables from objects which are in an array. But unfortunately it doesn't work, in fact, it doesn't output anything.
Therefore I tried to test it by replacing the function's body with just displaying the size of the array, but it prints 0 instead of 4 (the number of elements in that array).
Does that mean that the problem is within the function's parameters and how I pass them? Because on its own, ARRAY_SIZE works very well.
c++ arrays
asked Jan 18 at 14:19
John DavisJohn Davis
62
add a comment |
0
I have a problem with printing an array of struct objects.
Here is the code:
#include <iostream>
#include <string>
#define LOG(x) std::cout << x
#define ARR_SIZE(arr) sizeof(arr)/sizeof(arr[0])
struct vec2
{
int x, y;
};
struct Painting : vec2
{
std::string s_author, s_title, s_genre;
};
void disp_Array(Painting arr)
{
for (int i = 0; i < ARR_SIZE(arr); i++)
{
LOG(arr[i].s_author << "n");
LOG(arr[i].s_title << "n");
LOG(arr[i].s_genre << "n");
LOG(arr[i].x << "n");
LOG(arr[i].y << "n");
}
}
int main()
{
Painting painting[4];
disp_Array(painting);
for (int i = 0, j = 1; i < ARR_SIZE(painting), j <= ARR_SIZE(painting); i++, j++)
{
LOG("Painting " << j << "n");
LOG("Author: ");
std::cin.ignore();
std::getline(std::cin, painting[i].s_author);
LOG("Title: ");
std::getline(std::cin, painting[i].s_title);
LOG("Genre: ");
std::getline(std::cin, painting[i].s_genre);
LOG("Width: ");
std::cin >> painting[i].x;
LOG("Height: ");
std::cin >> painting[i].y;
LOG("n");
}
disp_Array(painting);
std::cin.get();
return 0;
}
The problem is within the disp_Array function. The idea was that the function should output the variables from objects which are in an array. But unfortunately it doesn't work, in fact, it doesn't output anything.
Therefore I tried to test it by replacing the function's body with just displaying the size of the array, but it prints 0 instead of 4 (the number of elements in that array).
Does that mean that the problem is within the function's parameters and how I pass them? Because on its own, ARRAY_SIZE works very well.
c++ arrays
asked Jan 18 at 14:19
John DavisJohn Davis
62
3
You can't copy C-style arrays, and thus can't pass them as parameters. Usestd::arrayorstd::vector, which both have asize()member function.
– Quentin
Jan 18 at 14:22
4
YourARR_SIZEonly work with real arrays. Once you pass the array to a function, it will decay to a pointer to its first element (as an argumentPainting arris equal toPainting* arr). And usingsizeofon a pointer gives the size of the pointer and not what it points to. My advice is to stop using old simple plain C-style arrays, and instead usestd::arrayorstd::vectorinstead.
– Some programmer dude
Jan 18 at 14:23
Thank you. I had that feeling that I should do it how it should be done and ease my life, but still I gave it a try. Also there is something strange, when I enter the second element's variables it skips Author. I fixed that with cin.ignore() but I should then press enter, how can I avoid pressing enter and still have that problem fixed?
– John Davis
Jan 18 at 14:49
@JohnDavis: it would be a different question but lies in the mixing ofgetline(line oriented) andcin>>(field oriented). After reading a field followed by enter, the end of line is still in input buffer and causes the nextgetlineto return an (almost) empty line. Long story short, dont mix.
– Serge Ballesta
Jan 18 at 14:54
add a comment |
0
0
0
I have a problem with printing an array of struct objects.
Here is the code:
#include <iostream>
#include <string>
#define LOG(x) std::cout << x
#define ARR_SIZE(arr) sizeof(arr)/sizeof(arr[0])
struct vec2
{
int x, y;
};
struct Painting : vec2
{
std::string s_author, s_title, s_genre;
};
void disp_Array(Painting arr)
{
for (int i = 0; i < ARR_SIZE(arr); i++)
{
LOG(arr[i].s_author << "n");
LOG(arr[i].s_title << "n");
LOG(arr[i].s_genre << "n");
LOG(arr[i].x << "n");
LOG(arr[i].y << "n");
}
}
int main()
{
Painting painting[4];
disp_Array(painting);
for (int i = 0, j = 1; i < ARR_SIZE(painting), j <= ARR_SIZE(painting); i++, j++)
{
LOG("Painting " << j << "n");
LOG("Author: ");
std::cin.ignore();
std::getline(std::cin, painting[i].s_author);
LOG("Title: ");
std::getline(std::cin, painting[i].s_title);
LOG("Genre: ");
std::getline(std::cin, painting[i].s_genre);
LOG("Width: ");
std::cin >> painting[i].x;
LOG("Height: ");
std::cin >> painting[i].y;
LOG("n");
}
disp_Array(painting);
std::cin.get();
return 0;
}
The problem is within the disp_Array function. The idea was that the function should output the variables from objects which are in an array. But unfortunately it doesn't work, in fact, it doesn't output anything.
Therefore I tried to test it by replacing the function's body with just displaying the size of the array, but it prints 0 instead of 4 (the number of elements in that array).
Does that mean that the problem is within the function's parameters and how I pass them? Because on its own, ARRAY_SIZE works very well.
c++ arrays
asked Jan 18 at 14:19
John DavisJohn Davis
62
I have a problem with printing an array of struct objects.
Here is the code:
#include <iostream>
#include <string>
#define LOG(x) std::cout << x
#define ARR_SIZE(arr) sizeof(arr)/sizeof(arr[0])
struct vec2
{
int x, y;
};
struct Painting : vec2
{
std::string s_author, s_title, s_genre;
};
void disp_Array(Painting arr)
{
for (int i = 0; i < ARR_SIZE(arr); i++)
{
LOG(arr[i].s_author << "n");
LOG(arr[i].s_title << "n");
LOG(arr[i].s_genre << "n");
LOG(arr[i].x << "n");
LOG(arr[i].y << "n");
}
}
int main()
{
Painting painting[4];
disp_Array(painting);
for (int i = 0, j = 1; i < ARR_SIZE(painting), j <= ARR_SIZE(painting); i++, j++)
{
LOG("Painting " << j << "n");
LOG("Author: ");
std::cin.ignore();
std::getline(std::cin, painting[i].s_author);
LOG("Title: ");
std::getline(std::cin, painting[i].s_title);
LOG("Genre: ");
std::getline(std::cin, painting[i].s_genre);
LOG("Width: ");
std::cin >> painting[i].x;
LOG("Height: ");
std::cin >> painting[i].y;
LOG("n");
}
disp_Array(painting);
std::cin.get();
return 0;
}
The problem is within the disp_Array function. The idea was that the function should output the variables from objects which are in an array. But unfortunately it doesn't work, in fact, it doesn't output anything.
Therefore I tried to test it by replacing the function's body with just displaying the size of the array, but it prints 0 instead of 4 (the number of elements in that array).
Does that mean that the problem is within the function's parameters and how I pass them? Because on its own, ARRAY_SIZE works very well.
c++ arrays
c++ arrays
asked Jan 18 at 14:19
John DavisJohn Davis
62
asked Jan 18 at 14:19
John DavisJohn Davis
62
asked Jan 18 at 14:19
John DavisJohn Davis
62
asked Jan 18 at 14:19
John DavisJohn Davis
62
asked Jan 18 at 14:19
John DavisJohn Davis
62
62
3
You can't copy C-style arrays, and thus can't pass them as parameters. Usestd::arrayorstd::vector, which both have asize()member function.
– Quentin
Jan 18 at 14:22
4
YourARR_SIZEonly work with real arrays. Once you pass the array to a function, it will decay to a pointer to its first element (as an argumentPainting arris equal toPainting* arr). And usingsizeofon a pointer gives the size of the pointer and not what it points to. My advice is to stop using old simple plain C-style arrays, and instead usestd::arrayorstd::vectorinstead.
– Some programmer dude
Jan 18 at 14:23
Thank you. I had that feeling that I should do it how it should be done and ease my life, but still I gave it a try. Also there is something strange, when I enter the second element's variables it skips Author. I fixed that with cin.ignore() but I should then press enter, how can I avoid pressing enter and still have that problem fixed?
– John Davis
Jan 18 at 14:49
@JohnDavis: it would be a different question but lies in the mixing ofgetline(line oriented) andcin>>(field oriented). After reading a field followed by enter, the end of line is still in input buffer and causes the nextgetlineto return an (almost) empty line. Long story short, dont mix.
– Serge Ballesta
Jan 18 at 14:54
add a comment |
3
You can't copy C-style arrays, and thus can't pass them as parameters. Usestd::arrayorstd::vector, which both have asize()member function.
– Quentin
Jan 18 at 14:22
4
YourARR_SIZEonly work with real arrays. Once you pass the array to a function, it will decay to a pointer to its first element (as an argumentPainting arris equal toPainting* arr). And usingsizeofon a pointer gives the size of the pointer and not what it points to. My advice is to stop using old simple plain C-style arrays, and instead usestd::arrayorstd::vectorinstead.
– Some programmer dude
Jan 18 at 14:23
Thank you. I had that feeling that I should do it how it should be done and ease my life, but still I gave it a try. Also there is something strange, when I enter the second element's variables it skips Author. I fixed that with cin.ignore() but I should then press enter, how can I avoid pressing enter and still have that problem fixed?
– John Davis
Jan 18 at 14:49
@JohnDavis: it would be a different question but lies in the mixing ofgetline(line oriented) andcin>>(field oriented). After reading a field followed by enter, the end of line is still in input buffer and causes the nextgetlineto return an (almost) empty line. Long story short, dont mix.
– Serge Ballesta
Jan 18 at 14:54
3
3
You can't copy C-style arrays, and thus can't pass them as parameters. Use
std::array or std::vector, which both have a size() member function.– Quentin
Jan 18 at 14:22
You can't copy C-style arrays, and thus can't pass them as parameters. Use
std::array or std::vector, which both have a size() member function.– Quentin
Jan 18 at 14:22
4
4
Your
ARR_SIZE only work with real arrays. Once you pass the array to a function, it will decay to a pointer to its first element (as an argument Painting arr is equal to Painting* arr). And using sizeof on a pointer gives the size of the pointer and not what it points to. My advice is to stop using old simple plain C-style arrays, and instead use std::array or std::vector instead.– Some programmer dude
Jan 18 at 14:23
Your
ARR_SIZE only work with real arrays. Once you pass the array to a function, it will decay to a pointer to its first element (as an argument Painting arr is equal to Painting* arr). And using sizeof on a pointer gives the size of the pointer and not what it points to. My advice is to stop using old simple plain C-style arrays, and instead use std::array or std::vector instead.– Some programmer dude
Jan 18 at 14:23
Thank you. I had that feeling that I should do it how it should be done and ease my life, but still I gave it a try. Also there is something strange, when I enter the second element's variables it skips Author. I fixed that with cin.ignore() but I should then press enter, how can I avoid pressing enter and still have that problem fixed?
– John Davis
Jan 18 at 14:49
Thank you. I had that feeling that I should do it how it should be done and ease my life, but still I gave it a try. Also there is something strange, when I enter the second element's variables it skips Author. I fixed that with cin.ignore() but I should then press enter, how can I avoid pressing enter and still have that problem fixed?
– John Davis
Jan 18 at 14:49
@JohnDavis: it would be a different question but lies in the mixing of
getline (line oriented) and cin>> (field oriented). After reading a field followed by enter, the end of line is still in input buffer and causes the next getline to return an (almost) empty line. Long story short, dont mix.– Serge Ballesta
Jan 18 at 14:54
@JohnDavis: it would be a different question but lies in the mixing of
getline (line oriented) and cin>> (field oriented). After reading a field followed by enter, the end of line is still in input buffer and causes the next getline to return an (almost) empty line. Long story short, dont mix.– Serge Ballesta
Jan 18 at 14:54
add a comment |
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3
You can't copy C-style arrays, and thus can't pass them as parameters. Use
std::arrayorstd::vector, which both have asize()member function.– Quentin
Jan 18 at 14:22
4
Your
ARR_SIZEonly work with real arrays. Once you pass the array to a function, it will decay to a pointer to its first element (as an argumentPainting arris equal toPainting* arr). And usingsizeofon a pointer gives the size of the pointer and not what it points to. My advice is to stop using old simple plain C-style arrays, and instead usestd::arrayorstd::vectorinstead.– Some programmer dude
Jan 18 at 14:23
Thank you. I had that feeling that I should do it how it should be done and ease my life, but still I gave it a try. Also there is something strange, when I enter the second element's variables it skips Author. I fixed that with cin.ignore() but I should then press enter, how can I avoid pressing enter and still have that problem fixed?
– John Davis
Jan 18 at 14:49
@JohnDavis: it would be a different question but lies in the mixing of
getline(line oriented) andcin>>(field oriented). After reading a field followed by enter, the end of line is still in input buffer and causes the nextgetlineto return an (almost) empty line. Long story short, dont mix.– Serge Ballesta
Jan 18 at 14:54