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POwhat does compiler do with a[i] which a is array? And what if a is a pointer?
 

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  1. POwhat does compiler do with a[i] which a is array? And what if a is a pointer?
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    <p>I was told by <a href="http://c-faq.com/aryptr/aryptr2.html" rel="nofollow noreferrer">c-faq</a> that compiler do different things to deal with a[i] while a is an array or a pointer. Here's an example from c-faq:</p> <blockquote> <pre><code>char a[] = "hello"; char *p = "world"; </code></pre> <p>Given the declarations above, when the compiler sees the expression a[3], it emits code to start at the location ``a'', move three past it, and fetch the character there. When it sees the expression p[3], it emits code to start at the location ``p'', fetch the pointer value there, add three to the pointer, and finally fetch the character pointed to.</p> </blockquote> <p>But I was told that when dealing with a[i], the compiler tends to convert a (which is an array) to a pointer-to-array. So I want to see assembly codes to find out which is right.</p> <p>EDIT:</p> <p>Here's the source of this statement. <a href="http://c-faq.com/aryptr/aryptrequiv.html" rel="nofollow noreferrer">c-faq</a> And note this sentence: </p> <blockquote> <p>an expression of the form a[i] causes the array to decay into a pointer, following the rule above, and then to be subscripted just as would be a pointer variable in the expression p[i] (although the eventual memory accesses will be different, " </p> </blockquote> <p>I'm pretty confused of this: since a has decayed to pointer, then why does he mean about "memory accesses will be different?"</p> <p>Here's my code:</p> <pre><code>// array.cpp #include &lt;cstdio&gt; using namespace std; int main() { char a[6] = "hello"; char *p = "world"; printf("%c\n", a[3]); printf("%c\n", p[3]); } </code></pre> <p>And here's part of the assembly code I got using g++ -S array.cpp</p> <pre><code> .file "array.cpp" .section .rodata .LC0: .string "world" .LC1: .string "%c\n" .text .globl main .type main, @function main: .LFB2: leal 4(%esp), %ecx .LCFI0: andl $-16, %esp pushl -4(%ecx) .LCFI1: pushl %ebp .LCFI2: movl %esp, %ebp .LCFI3: pushl %ecx .LCFI4: subl $36, %esp .LCFI5: movl $1819043176, -14(%ebp) movw $111, -10(%ebp) movl $.LC0, -8(%ebp) movzbl -11(%ebp), %eax movsbl %al,%eax movl %eax, 4(%esp) movl $.LC1, (%esp) call printf movl -8(%ebp), %eax addl $3, %eax movzbl (%eax), %eax movsbl %al,%eax movl %eax, 4(%esp) movl $.LC1, (%esp) call printf movl $0, %eax addl $36, %esp popl %ecx popl %ebp leal -4(%ecx), %esp ret </code></pre> <p>I can not figure out the mechanism of a[3] and p[3] from codes above. Such as:</p> <ul> <li>where was "hello" initialized? </li> <li>what does $1819043176 mean? maybe it's the memory address of "hello" (address of a)?</li> <li>I'm sure that "-11(%ebp)" means a[3], but why?</li> <li>In "movl -8(%ebp), %eax", content of poniter p is stored in EAX, right? So $.LC0 means content of pointer p?</li> <li>What does "movsbl %al,%eax" mean? </li> <li><p>And, note these 3 lines of codes:<br> movl $1819043176, -14(%ebp)<br> movw $111, -10(%ebp)<br> movl $.LC0, -8(%ebp) </p> <p>The last one use "movl" but why did not it overwrite the content of -10(%ebp)? (I know the anser now :), the address is incremental and "movl $.LC0 -8(%ebp) will only overwrite {-8, -7, -6, -5}(%ebp))</p></li> </ul> <p>I'm sorry but I'm totally confused of the mechanism, as well as assembly code... </p> <p>Thank you very much for your help.</p>
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    1. COI think your this statement "the compiler tends to convert a (which is an array) to a pointer-to-array" is not correct. Please tell me who said this to you?
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    2. CO+1, for trying it out yourself before asking.
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    3. CONot pointer-to-array, pointer-to-char.
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