Microsoft KB Archive/44463

= How To Difference Between Arrays and Pointers in C =

Article ID: 44463

Article Last Modified on 7/5/2005

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APPLIES TO


 * Microsoft C Professional Development System 6.0a
 * Microsoft Visual C++ 1.0 Professional Edition
 * Microsoft Visual C++ 1.5 Professional Edition
 * Microsoft Visual C++ 1.51
 * Microsoft Visual C++ 1.52 Professional Edition
 * Microsoft Visual C++ 2.0 Professional Edition
 * Microsoft Visual C++ 2.1
 * Microsoft Visual C++ 4.0 Standard Edition
 * Microsoft Visual C++ 5.0 Enterprise Edition
 * Microsoft Visual C++ 6.0 Enterprise Edition
 * Microsoft Visual C++ 5.0 Professional Edition
 * Microsoft Visual C++ 6.0 Professional Edition
 * Microsoft Visual C++ 6.0 Standard Edition

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This article was previously published under Q44463





SUMMARY
The text below presents an example of a common programming mistake, that is, confusing an array and a pointer declaration:

Consider an application divided into several modules. In one module, declare an array as follows:

signed char buffer[100]; In another module, declare the following variables to access the array: extern signed char *buffer;          // FAILS extern signed char buffer[];         // WORKS If you view the code in the CodeView debugger, it indicates that the &quot;*buffer&quot; declaration produces a different address than the &quot;buffer[]&quot; declaration does.



MORE INFORMATION
The following declarations are NOT the same: char *pc; char ac[20]; The first declaration allocates memory for a pointer; the second allocates memory for 20 characters.

A picture of pc and ac in memory might appear as follows:    pc  ++ |  ???  |        ++

ac +-+-+-+-+     +-+ | ?  |  ?  |  ?  |  ?  | ... |  ?  |        +-+-+-+-+     +-+ Neither are the following the same: extern char *pc; extern char ac[]; The first declaration indicates that another module allocated either two or four bytes for a pointer to char named pc while the second indicates that another module allocated an array (of some unspecified length) named ac.

The steps required to address pc[3] and ac[3] are different. The one similarity is that the expression &quot;ac&quot; is a constant pointer to char that points to &ac[0].

To evaluate pc[3], load the value of the pointer pc from memory and add 3. Then load the character stored ad this location (pc + 3) into a register. Assuming the small memory model, the appropriate MASM code might look like the following: MOV    BX, pc          ; move *CONTENTS* of pc into BX                           ; BX contains 1234 MOV    AL, [BX + 3]    ; move byte at pc + 3 (1237) into AL                           ; ==> AL contains 'd' An appropriate diagram might appear as follows, provided that pc has been set to point to an array at location 1234 and that the first four positions of the array contain the string &quot;abcd&quot;: address:  1000                  1234  1235  1236  1237 pc ++--->>>>>--v-v-v-v-+ | 1234  |          *pc |  a  |  b  |  c  |  d  | ... ++             +-+-+-+-+                                 pc[0] pc[1] pc[2] pc[3] *pc  *(pc+1) and so on NOTE: If you use pc without first initializing it properly causes your application to access random memory which can cause undesired behavior. To initialize the pointer, include a line of code such as &quot;pc = malloc(5);&quot; or &quot;pc = ac;&quot;.

Because ac is a constant, it can be stored in the final MOV instruction, which eliminates two MOV instructions. The MASM code might look like the following: MOV    AL, [offset ac + 3]     ; move byte at ac + 3 into AL                                   ; offset ac is 1100, so move ; byte at 1103 into AL                                  ; ==> AL contains 'd' The corresponding picture might appear as follows: address: 1100 1101  1102  1103        1119 ac +-+-+-+-+     +-+ | a  |  b  |  c  |  d  | ... | \0 |        +-+-+-+-+     +-+        ac[0] ac[1] ac[2] ac[3]       ac[19] *ac  *(ac+1)  and so on NOTE: If you first initialize pc to point to ac (by including the line &quot;pc = ac;&quot; in your application), then the end result of the two statements is identical. To see this in the picture, set pc to contain the address of ac, 1100. However, the instructions used to generate these effects are quite different.

If you declare ac as follows, the compiler generates code to perform pointer-type addressing rather than array-type addressing: extern char *ac; // WRONG! The compiler uses the first few bytes of the array as an address (rather than as characters) and accesses the memory stored at that (unintended) location.

Additional query words: 8.00

Keywords: kbhowto kblangc KB44463

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