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copied!<ul> <li><code>int *a;</code> This defines a variable which is a pointer to an integer type. The pointer type variable <code>a</code> at the creation contains garbage value.</li> <li>When you do <code>*a = 10;</code> it ties to use the value stored in <code>a</code> , which is garbage, as an address and store the value <code>10</code> there. Because we do not know what <code>a</code> contains and it is not allocated so <code>a</code> points to some memory location which is unknown and accessing it will be illegal, and will get you a segmentation fault (or something similar).</li> <li>Same in the case of <code>printf ("%d", *a);</code> . This also tries to access the value stored at some undefined memory location which you have not allocated.</li> </ul> <p>.</p> <pre><code>this variable is this is the location stored in some with the address address on the 'garbage'. You do not stack have permissions to access this +-------+---------+ | name | value | +-------+---------+ +---------+ | a | garbage |---->| ????? | +-------+---------+ +---------+ </code></pre> <p>After you have defined the pointer type variable, you need to request for some memory location from the operating system and use that memory location value to store into <code>a</code>, and then use that memory location through <code>a</code>.</p> <p>To do that you need to do the following:</p> <pre><code>int *a; a = malloc (sizeof (int)); /* allocates a block of memory * of size of one integer */ *a = 10; printf ("%d", *a); free (a); /* You need to free it after you have used the memory * location back to the OS yourself. */ </code></pre> <p>In this case it is like below:</p> <p>After <code>*a = 10;</code> . The pointer variable is allocated in the stack. At this moment the <code>a</code> contains a garbage value. Then <code>a</code> points to an address with that garbage value.</p> <pre><code>this variable is this is the location stored in some with the address address on the 'garbage'. You do not stack have permissions to access this +-------+---------+ | name | value | +-------+---------+ +---------+ | a | garbage |---->| ????? | +-------+---------+ +---------+ </code></pre> <p>After <code>a = (int *) malloc (sizeof (int));</code> . Let us assume that <code>malloc</code> returns you some address <code>0x1234abcd</code>, to be used. At this moment <code>a</code> will contain <code>0x1234abcd</code> then <code>a</code> points to a valid memory location which was allocated and reserved for you to be used. But note that the value inside <code>0x1234abcd</code> can be anything, ie. garbage. You can use <code>calloc</code> to set the contents of the memory locations you allocate to <code>0</code>.</p> <pre><code>this variable is this is the location stored in some 0x1234abcd , allocated address on the by malloc, and reserved stack for your program. You have access to this location. +-------+------------+ | name | value | +-------+------------+ +---------+ | a | 0x1234abcd |---->| garbage| +-------+------------+ +---------+ </code></pre> <p>After <code>*a = 10;</code> , by <code>*a</code> you access the memory location <code>0x1234abcd</code> and store <code>10</code> into it.</p> <pre><code>this variable is this is the location stored in some 0x1234abcd , allocated address on the by malloc, and reserved stack for your program. You have access to this location. +-------+------------+ | name | value | +-------+------------+ +---------+ | a | 0x1234abcd |---->| 10 | +-------+------------+ +---------+ </code></pre> <p>After <code>free (a)</code> , the contents of <code>a</code> ie. the memory address <code>0x1234abcd</code> will be freed, ie returned back to the operating system. Note that after freeing the <code>0x1234abcd</code> the contents of <code>a</code> is <em>still</em> <code>0x1234abcd</code> , but you can no more access it legally, because you just freed it. Accessing the contents pointed by the address stored in <code>a</code> will result in undefined behavior, most probably a segmentation fault or heap corruption, as it is freed and you do not have access rights.</p> <pre><code>this variable is this is the location stored in some 0x1234abcd , allocated address on the by malloc. You have freed it. stack Now you CANNOT access it legally +-------+------------+ | name | value | +-------+------------+ +---------+ | a | 0x1234abcd | | 10 | +-------+------------+ +---------+ the contents of a remains the same. </code></pre> <p><strong>EDIT1</strong></p> <p>Also note the difference between <code>printf ("%d", a);</code> and <code>printf ("%d", *a);</code> . When you refer to <code>a</code> , it simply prints the contents of <code>a</code> that is <code>0x1234abcd</code>. And when you refer <code>*a</code> then it uses <code>0x1234abcd</code> as an address , and then prints the contents of the address, which is <code>10</code> in this case.</p> <pre><code>this variable is this is the location stored in some 0x1234abcd , allocated address on the by malloc, and reserved stack for your program. You have access to this location. +-------+------------+ | name | value | +-------+------------+ +---------+ | a | 0x1234abcd |---->| 10 | +-------+------------+ +---------+ ^ ^ | | | | (contents of 'a') (contents of the ) | (location, pointed ) printf ("%d", a); ( by 'a' ) | +----------------+ | printf ("%d", *a); </code></pre> <p><strong>EDIT2</strong></p> <p>Also note that <code>malloc</code> can fail to get you some valid memory location. You should always check if <code>malloc</code> returned you a valid memory location. If <code>malloc</code> cannot get you some memory location to be used then it will return you <code>NULL</code> so you should check if the returned value is <code>NULL</code> or not before use. So finally the code becomes:</p> <pre><code>int *a; a = malloc (sizeof (int)); /* allocates a block of memory * of size of one integer */ if (a == NULL) { printf ("\nCannot allocate memory. Terminating"); exit (1); } *a = 10; printf ("%d", *a); free (a); /* You need to free it after you have used the memory * location back to the OS yourself. */ </code></pre>

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