I want to set all the index value to -1 in a double array.
Here is my code :
double dp[505]; memset(dp,-1,sizeof(dp)); cout<<dp[0]<<"\n"; But it is showing nan when i try to print its value.
What does nan mean? Is it possible to use memset() in double array?
In C++, you can write:
double initValue = -1; std::fill_n(dp, 505, initValue); memsetting a double array with a non-double value won't work.
memset operates on bytes, not floats, and a double with all bytes set to -1 does not equal -1. I think you're looking for std::fill:
#include <algorithm> std::fill(dp, dp + 505, -1.0); Or, in C++11:
#include <algorithm> #include <iterator> std::fill(std::begin(dp), std::end(dp), -1.0); From the man page of memset:
The
memset()function fills the first n bytes of the memory area pointed to by s with the constant byte c.
The problem is that you want to fill an array of doubles with the constant -1.0 but sizeof(double) > 1 so memset actually fills in garbage which happens to end up as a NaN.
If you are using C++, the std::fill function is your friend. Actually, since you are writing to your array for the first time, std::uninitialized_fill would be correct. Although for the builtin double type there should be no difference but it is always good to be precise.
constexpr std::size_t length = 505; double values[length]; std::uninitialized_fill(values, values + length, -1.0); You have set each element of the array to be filled with the byte 0xFF (i.e. the char representation of -1).
No floating point number is represented by a series of 0xFF bytes, so on printing the double, you see NaN (i.e. 'not a number'). This is in apparent contrast to memset'ting the bytes to zero, which is legal as a string of 0 bytes is a double with value zero. See Is it legal to use memset(,0,) on array of doubles?.
If you meant to set every entry to -1.0 (i.e. a double), then use std::fill or std::fill_n in C++ or a loop in C, e.g.
int n; for (n = 0 ; n < 505 ; n++) dp[n] = -1.0; memset sets bytes, so you get double-values where each byte is -1.
Instead in C++ use std::vector, then write
vector<double> dp( 505, -1.0 ); It's that simple.
If dp is a global vector and you need to set it to -1 a number of times, then you can simply do this:
dp = vector<double>( dp.size(), -1.0 ); However, it's generally not a good idea to use non-const global variables.
Alternatively one can use std::fill, or just a loop, or just about any technique that still treat the double values as double values. But std::vector is preferable also for many other reasons than greatly simplifying the fill-it task. In particular a std::vector can be resized, it takes care of copying, and it automates the memory management, doing that part correctly and transparent to you.
nan means not a number.
cant see why its not working. maybe because precision is not set : (cout.precision(15);)
check this: How do I print a double value with full precision using cout?
But im not sure at all it will works :o i checked memset source code and there's no problem with negative :D
But it can be a problem with doubles :
memset(dst0, c0, length) void *dst0; register int c0; register size_t length; Have you tried to compile with Werror flag ?
Although answers for you question have been given, I just wanted you to note that sizeof(dp) outputs the number of bytes used to code the variable in memory.
In your case, dp is a pointer to a double. It will then be equal to the size of a pointer (4), no matter wether or not memery has been allocated. sizeof(*dp) will output the size of a double (8). In order to use the length of a