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Add some explanations of how apint loads and stores

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work.

llvm-svn: 67471
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Duncan Sands 2009-03-22 11:33:16 +00:00
parent 7299f4d1e4
commit dca010d0d7
1 changed files with 15 additions and 2 deletions
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docs/LangRef.html
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@ -3486,7 +3486,13 @@ alignment may produce less efficient code. An alignment of 1 is always
safe.
</p>
<h5>Semantics:</h5>
<p>The location of memory pointed to is loaded.</p>
<p>The location of memory pointed to is loaded. If the value being loaded
is of scalar type then the number of bytes read does not exceed the minimum
number of bytes needed to hold all bits of the type. For example, loading an
<tt>i24</tt> reads at most three bytes. When loading a value of a type like
<tt>i20</tt> with a size that is not an integral number of bytes, the result
is undefined if the value was not originally written using a store of the
same type.</p>
<h5>Examples:</h5>
<pre> %ptr = <a href="#i_alloca">alloca</a> i32 <i>; yields {i32*}:ptr</i>
<a
@ -3525,7 +3531,14 @@ safe.
</p>
<h5>Semantics:</h5>
<p>The contents of memory are updated to contain '<tt>&lt;value&gt;</tt>'
at the location specified by the '<tt>&lt;pointer&gt;</tt>' operand.</p>
at the location specified by the '<tt>&lt;pointer&gt;</tt>' operand.
If '<tt>&lt;value&gt;</tt>' is of scalar type then the number of bytes
written does not exceed the minimum number of bytes needed to hold all
bits of the type. For example, storing an <tt>i24</tt> writes at most
three bytes. When writing a value of a type like <tt>i20</tt> with a
size that is not an integral number of bytes, it is unspecified what
happens to the extra bits that do not belong to the type, but they will
typically be overwritten.</p>
<h5>Example:</h5>
<pre> %ptr = <a href="#i_alloca">alloca</a> i32 <i>; yields {i32*}:ptr</i>
store i32 3, i32* %ptr <i>; yields {void}</i>