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copied!<p>C++ adopts the C syntax. As revealed in "<a href="http://cm.bell-labs.com/cm/cs/who/dmr/chist.html" rel="nofollow noreferrer">The Development of the C Language</a>" (by Dennis Ritchie) C uses <code>*</code> for pointers in type declarations because it was decided that type syntax should follow use.</p> <blockquote> <p>For each object of [a compound type], there was already a way to mention the underlying object: index the array, call the function, use the indirection operator [<code>*</code>] on the pointer. Analogical reasoning led to a declaration syntax for names mirroring that of the expression syntax in which the names typically appear. Thus,</p> <pre><code>int i, *pi, **ppi; </code></pre> <p>declare an integer, a pointer to an integer, a pointer to a pointer to an integer. The syntax of these declarations reflects the observation that i, *pi, and **ppi all yield an int type when used in an expression. </blockquote></p> <p>Here's a more complex example:</p> <pre><code>int *(*foo)[4][]; </code></pre> <p>This declaration means an expression <code>*(*foo)[4][0]</code> has type <code>int</code>, and from that (and that <code>[]</code> has higher precedence than unary <code>*</code>) you can decode the type: foo is a pointer to an array of size 4 of array of pointers to ints.</p> <p>This syntax was adopted in C++ for compatibility with C. Also, don't forget that C++ has a use for & in declarations.</p> <pre><code>int & b = a; </code></pre> <p>The above line means a reference variable refering to another variable of type <code>int</code>. The difference between a reference and pointer roughly is that references are initialized only, and you can not change where they point, and finally they are always dereferenced automatically.</p> <pre><code>int x = 5, y = 10; int& r = x; int sum = r + y; // you do not need to say '*r' automatically dereferenced. r = y; // WRONG, 'r' can only have one thing pointing at during its life, only at its infancy ;) </code></pre>

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