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Because the type of the accumulator that will be used to determine the sum is based on the type of the third input argument. The literal 0 is not being interpreted as long long int. If you cast the initial value to accumulate appropriately, you get the right answer.
accumulate(arr.begin(),arr.end(),static_cast<long long int>(0))
I think there is another way , to avoid the static_cast(0) , and do it by simply accumulate(....,0L);
If I remember correct (I am too lazy to search in books or google now to verify!) the L forces the number to be interpreted as long long.
Hi, I understand the 3rd argument of the function accumulate(), which indicates the type of integer literal in this case. But why we need to cast statically as
A Very Big Sum
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Because the type of the accumulator that will be used to determine the sum is based on the type of the third input argument. The literal 0 is not being interpreted as long long int. If you cast the initial value to accumulate appropriately, you get the right answer.
From http://stackoverflow.com/questions/8108642/type-of-integer-literals-not-int-by-default
"The type of an integer literal is the first of the corresponding list in Table 6 in which its value can be represented."
And Table 6, for literals without suffixes and decimal constants, gives:
So, without the cast, the literal 0 can be represented as int, so accumulate uses int as the type for accumulation.
I think there is another way , to avoid the static_cast(0) , and do it by simply accumulate(....,0L); If I remember correct (I am too lazy to search in books or google now to verify!) the L forces the number to be interpreted as long long.
L is the suffix for long just like LL is the suffix for long long. Weird but not confusing.
accumulate(....,0LL)avoids the static cast.or we can do it more simply like this..
accumulate(arr.begin, arr.end(), (long long int)0);
Hi, I understand the 3rd argument of the function accumulate(), which indicates the type of integer literal in this case. But why we need to cast statically as
for the long long integer?
i also don't know that why i cast it explicitly
I am having trouble grasping a clear definition of initial value. What would that mean?