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copied!<p>When you use the <code>%s</code> format specifier, it expect a value which is a pointer to the start of a string. In C, this type is <code>char *</code>.</p> <p>Taking your <code>main</code> function, your variable <code>s1</code> is of type <code>char *</code>. Therefore, <code>s1</code> is a valid parameter to <code>printf</code>, so this line is valid:</p> <pre><code>printf("%s\n", s1); </code></pre> <p>Note the <em>absence</em> of an <code>&</code> in front of <code>s1</code>. In your code, you used the <code>&</code>, which takes the address of <code>s1</code>, the result of which will be of type <code>char **</code>. This is the <em>wrong</em> type, so <em>don't</em> use the <code>&</code>.</p> <p>The thing is, <code>printf</code> can't actually tell what type its arguments are, since it is a variadic function. It simply uses whatever arguments are there, according to the types specified in the format string.</p> <p>The same thing goes for <code>scanf</code>, but there is a pitfall: you must make sure that enough memory is allocated to account for the user input, else you will experience a buffer overflow with unpredictable results. Aside from this, <code>printf</code> and <code>scanf</code> are perfectly complementary.</p> <p>Anyhoo, this takes care of the compiler warnings, aside from the unused <code>cmds</code> variable (it's unnecessary in the provided code). Also, there is the part of <code>args</code> - it really should be a variable declared inside of <code>shell</code>, and not passed as a parameter, since its value is not even used inside <code>shell</code>.</p> <p>Don't know what's up with the rest of the code. It's superfluous considering your <code>main</code> function only calls on <code>shell</code>.</p>

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