This expr ??

[MaXiMuS]

Hi all,

-->> size *= (int)dims[j];

-->> retVal.m_vDims.push_back( dims[j] );

what do these two line of code mean? size is of type int.

thnks

 

[Klaas Vantournhout]

Hi all,

-->> size *= (int)dims[j];

-->> retVal.m_vDims.push_back( dims[j] );

what do these two line of code mean? size is of type int.

thnks

This depends on what retVal, n_vDims and dims are.

 

[Ian Collins]

Hi all,

-->> size *= (int)dims[j];

-->> retVal.m_vDims.push_back( dims[j] );

what do these two line of code mean? size is of type int.

You jest? Do you have a C++ book?

 

[Jonathan Lane]

Hi all,

-->> size *= (int)dims[j];

this calls operator *= on size. Same as writing
size.operator*=( (int)dims[j]);

-->> retVal.m_vDims.push_back( dims[j] );

This calls the push_back function on a member of class retVal called
m_vDims. It passes in the result of calling operator[] on dims with a
parameter of j.

 

[Jim Langston]

Hi all,

-->> size *= (int)dims[j];

*= is times equals.
a *= b;
is same as
a = a * b;
so this is
size = size * (int)dims[j];

If you can't figure it out from there, then you need to start back over from
page 1 of your book.

-->> retVal.m_vDims.push_back( dims[j] );

retVal is a structure or class.
It has some class/containter called m_vDims. maybe a vector.
it calls the functoin push_back of this instance. If it is a vector it is
pushing the value of dims[j] onto the vector at the end.

Start on page one of your book please.