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a = (int*)malloc(sizeof(int)*N); 
b = (int*)malloc(sizeof(int)*N);

If I have allocated some memory, how do I initialize both a and b to 1's ?

Adrian Mole
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Musab Ahmad
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  • Casting the result of `malloc()` is frowned upon ... https://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc – pmg Jun 13 '20 at 11:56
  • @pmg I was just thinking of that myself (see edited answer) - while I was taking a bath! I normally point it out (and have done in most of my other relevant posts). – Adrian Mole Jun 13 '20 at 12:10
  • As you have the `cuda` tag, maybe worth considering `cuMemAlloc()`, in which case you can then use `cuMemsetD32()` on the memory block. – Adrian Mole Jun 13 '20 at 12:30
  • @pmg if you want your code to compile with a C++ compiler, a cast is needed – VMMF Jul 29 '22 at 15:58

1 Answers1

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Unfortunately, there is no standard way of initializing a memory block (of non-byte-size data) to anything other than all zeros without using a loop. (For all zeros, you can use the calloc function; and, for a block of single-byte values, you can use 'memset()')

In your case, as the two blocks are the same size, and you want to initialize both to the same values, you can use a simple, one-line loop:

a = malloc(sizeof(int)*N); // No need to cast the return value of malloc!
b = malloc(sizeof(int)*N);
// Initialize:
for (int i = 0; i < N; ++i) a[i] = b[i] = 1;

There are various 'tricks' you might consider to optimize the for loop, but such practices are risky, platform-specific hacks, and often achieve very little in terms of execution speed. Further, most modern compilers will make short-shrift of that loop, using techniques such as loop unrolling, vectorization and CPU-specific intrinsics (MSVC, for example, generates code to use a 64-bit integer with a value of 0x0000000100000001 to cut the number of loops in half).

On not casting the result of malloc, see: Do I cast the result of malloc?
Adrian Mole
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