I have a vector int a[100], initialised with values such as 9726745. How do I set them all to a specific value x without a for statement?
I have a vector int a[100], initialised with values such as 9726745. How do I set them all to a specific value x without a for statement?
I don't think you can do it without looping (at least indirectly).
The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.
>without a for statement?
With a while statement
You can use memset() to do this.
Edited: Irrelevant.
Last edited by XSquared; 07-01-2003 at 02:15 PM.
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Never mind the fact that there are no vectors in C.Originally posted by XSquared
It's not safe to use memset( ) with a vector.
Quzah.
Hope is the first step on the road to disappointment.
Are we talking about an array here or a vector class? For an array, my example will most certainly work. For a vector class it won't, but this is the C programming board not C++.
Heh. Oops.
As soon as I saw vector my mind snapped into C++ mode.
Naturally I didn't feel inspired enough to read all the links for you, since I already slaved away for long hours under a blistering sun pressing the search button after typing four whole words! - Quzah
You. Fetch me my copy of the Wall Street Journal. You two, fight to the death - Stewie
Proper handling of memset is save with an array.
It is always save to set all bits to zero in an array of char (signed or unsigned),but thats all.C allows every single
data type except {signed,unsigned} char to have padding bits, that are not part
of the representation of the value. It is possible that improper
combinations of these padding bits can generate what is called a trap
representation.
So,for example memseting an array of ints to zero is not save (nonportable).
[OT]
The memory of std::vector is continous,memseting a std::vector of {unsigned,signed} chars is save too.
[/OT]
Please, just take the easy obvious way:
Code:int a[100]; for (int i=0;i<100;i++) { a[i]=0; }
Last edited by VirtualAce; 07-02-2003 at 06:37 AM.
One way to assign to an array without a loop is to wrap the array in a structure.Code:#include <stdio.h> struct sType { int a[10]; }; const struct sType Default = {{56,7,8,1,27,13,54,82,23,16}}; void show(const int *array, size_t size) { size_t i; for ( i = 0; i < size; ++i ) { printf("%2d%c", *array++, i % 10 == 9 ? '\n' : ','); } } int main(void) { struct sType object = {{83,6,71/*remaining elements are 0*/}}; fputs("Before: ", stdout); show(object.a, sizeof(object.a)/sizeof(*object.a)); object = Default; fputs("After: ", stdout); show(object.a, sizeof(object.a)/sizeof(*object.a)); return 0; } /* my output Before: 83, 6,71, 0, 0, 0, 0, 0, 0, 0 After: 56, 7, 8, 1,27,13,54,82,23,16 */
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
Here's an obvious answer: at the time of declaration:
int foo[5] = { 1, 1, 1, 1, 1 };
Of course, you could always use recursion:
There's a few missed ways. That'll suffice for now.Code:void fill( int array[], size_t size, int value ) { if( size > -1 ) { fill( array, size-1, value ); array[size] = value; } }
Quzah.
Hope is the first step on the road to disappointment.
