How to initialize a 2D array with pointer.
int *mapTable[] = { {1,10} , {2,20} , {3,30} , {4,40} }; // It's giving error
Here int *mapTable is giving an error.
How can I declare them properly?
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1Although I can tell what is happening, in general, please don't say "it's giving error" but tell us what that error is.Lightness Races in Orbit– Lightness Races in Orbit2011-06-24 09:01:05 +00:00Commented Jun 24, 2011 at 9:01
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2 Answers
int *mapTable[] is not a 2D array: it is a 1D array of pointers.
But then you go and use a 2D array initialiser { {1,10} , {2,20} , {3,30} , {4,40} }.
That's why it's "giving error".
The 2D array way
Try:
int mapTable[][2] = { {1,10} , {2,20} , {3,30} , {4,40} };
And, yes, you do need to specify the size of that second dimension.
The 1D array of pointers way
This is a little more involved, and is usually too complex to be worth it.
It also usually requires dynamic allocation, causing a total mess with object lifetime:
int *mapTable[] = { new int[2], new int[2], new int[2], new int[2] };
int main() {
mapTable[0][0] = 1; mapTable[0][1] = 10;
mapTable[1][0] = 2; mapTable[1][1] = 20;
mapTable[2][0] = 3; mapTable[2][1] = 30;
mapTable[3][0] = 4; mapTable[3][1] = 40;
// then, at the end of your program:
for (size_t i = 0; i < 4; i++)
delete[] mapTable[i];
}
As you can see, this is not ideal.
You can avoid dynamic allocation:
int mapTable0[] = {1,10};
int mapTable1[] = {2,20};
int mapTable2[] = {3,30};
int mapTable3[] = {4,40};
int *mapTable[] = { &mapTable0[0], &mapTable1[0], &mapTable2[0], &mapTable3[0] };
But I'm not sure why you'd want to go down this avenue.
answered Jun 24, 2011 at 9:02
Lightness Races in Orbit
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5 Comments
Lightness Races in Orbit
@Cicada: Arrays are not pointers. Please stop spreading this dangerous misinformation.
Paweł Sołtysiak
Arrays are pointers! But have const nature, you can't move them
Lightness Races in Orbit
@Soltys: No, they completely are not. This is utter nonsense.
Jerry Coffin
@Soltys: no, they're not. The name of an array, under the right circumstances, can decay to an rvalue that's of the right type to be assigned to a pointer -- but even then, the array definitely is not a pointer. Calling it a pointer is like calling
1 an int variable. It's a value that can be assigned to an int variable, but it's not an int variable itself at all.
sbi
@Soltys: How do I use arrays in C++?.
The questin is tagged C++ and C; but all these answers only cover C++ (for the pointer way).
So in C, you can declare the rows separately and then compose them.
int r0[] = {1,10}, r1[] = {2,20}, r2[] = {3,30}, r3[] = {4,40};
int *mapTable[] = { r0, r1, r2, r3 };
Or, using C99, you can make anonymous arrays in the initializer.
int *mapTable[] = { (int[]){1,10}, (int[]){2,20}, (int[]){3,30}, (int[]){4,40} };
Both of these use the fact that an array reference decays into a pointer.
