How to copy and assign an array of arrays?

Question

  int zero[5][4] = {
    { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }
  };
  int m1[5][4] = {
    { 1, 0, 0, 0 }, { 0, 1, 0, 0 }, { 0, 0, 1, 0 }, { 0, 0, 0, 1 }
  };

  //errors here
  m1 = zero;
  m1 = { { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 } };
  m1[0] = { 0, 0, 0, 0 };

Is there no syntax for this? Do I have to use a loop with indexing to achieve this?

Is memcpy suitable for you? or memset if you want to set to all zero. — kaylum
– kaylum, Commented Dec 15, 2015 at 0:09

haccks · Accepted Answer · 2015-12-15 00:41:24Z

4

In C, arrays are not assignable or can't be on the left side of assignment (=) operator. Use memcpy.

memcpy(m1, zero, sizeof(zero)); // Assuming size of 'm1' is no less than the size of 'zero'

You can use memset to set an array with a constant value like 0 or -1

memset(m1, 0, sizeof(m1));

edited Dec 15, 2015 at 0:41

answered Dec 15, 2015 at 0:10

haccks

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2 Comments

too honest for this site Over a year ago

port70.net/~nsz/c/c11/n1570.html#7.24.6.1p2 memset takes basically an unsigned char, so you must not use -1. Also one should clarify that any other value than 0 is implementation defined - at best, because memset does not care about the datatype of the target. It just writes bytes.

too honest for this site Over a year ago

I don't see the point. It still is implementation defined for negative values. And there already have been quite some questions/problems on SO where the poster thought memset would know the type of the array (the problem arises less for structs). My claim stands: except for char [], memset is only useful to zero an array and even that is questionable for types other than integers, because only for these "all bits 0" really means the value 0. There is no requirement for e.g. a null pointer to have all bits cleared, nor for a float to interpret that as 0.0.

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