Yes, it can apply to any memory buffer, but you must input the correct memory buffer size memset treats any memory buffer as a series of bytes, so whether it's char int float double etc, doesn't really matter. Keep in mind though that it will not set multi-byte types to a specific non-zero value ... for example.
Yes, it can apply to any memory buffer, but you must input the correct memory buffer size ... memset treats any memory buffer as a series of bytes, so whether it's char, int, float, double, etc, doesn't really matter. Keep in mind though that it will not set multi-byte types to a specific non-zero value ... for example: int a100; memset(a, 1, sizeof(a)); will not set each member of a to the value 1 ... rather it will set every byte in the memory buffer taken up by a to 1, which means every four-byte int will be set to the value 0x01010101, which is not the same as 0x00000001.
For static-sized and variable-length arrays, you can just foo ...; memset (foo, 0, sizeof (foo)); // sizeof() gives size of entity in bytes Rule of thumb: Never hardcode data sizes. (This does not work if you pass arrays as function arguments: Behaviour of Sizeof in C ).
It can be applied to any array. The 100 at the end is the size in bytes, so a integer would be 4 bytes each, so it would be - int a100; memset(a, 0, sizeof(a)); //sizeof(a) equals 400 bytes in this instance Get it? :).
PS: It's not just arrays, its arbitrary memory blocks, you're setting memory from location to value 0 for a length of 400 bytes in my example - it doesn't care what the memory was formatted as. – w00te Jul 25 '11 at 12:56 arbitray memory blocks? – ratzip Jul 25 '11 at 12:58 1 -1: Never hardcode type sizes.
– phresnel Jul 25 '11 at 13:00 you really need to mention sizeof – David Heffernan Jul 25 '11 at 13:01.
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