Article: Q44463
Product(s): Microsoft C Compiler
Version(s): 1.0,1.5,1.51,1.52,2.0,2.1,4.0,5.0,6.0
Operating System(s):
Keyword(s): kbLangC kbVC100 kbVC150 kbVC151 kbVC152 kbVC200 kbVC210 kbVC400 kbVC500 kbVC600
Last Modified: 28-SEP-2001
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The information in this article applies to:
- Microsoft C for MS-DOS
- Microsoft C/C++ for MS-DOS
- Microsoft Visual C++, versions 1.0, 1.5, 1.51, 1.52, 2.0, 2.1, 4.0
- Microsoft Visual C++, 32-bit Enterprise Edition, versions 5.0, 6.0
- Microsoft Visual C++, 32-bit Professional Edition, versions 5.0, 6.0
- Microsoft Visual C++, 32-bit Learning Edition, version 6.0
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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 "*buffer"
declaration produces a different address than the "buffer[]" 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 "ac" 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 "abcd":
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 "pc = malloc(5);" or "pc
= ac;".
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 "pc = ac;"
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
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Keywords : kbLangC kbVC100 kbVC150 kbVC151 kbVC152 kbVC200 kbVC210 kbVC400 kbVC500 kbVC600
Technology : kbVCsearch kbVC400 kbAudDeveloper kbZNotKeyword8 kbvc150 kbvc100 kbCCompSearch kbZNotKeyword3 kbVC500 kbVC600 kbVC151 kbVC200 kbVC210 kbVC32bitSearch kbVC152 kbVC500Search
Version : :1.0,1.5,1.51,1.52,2.0,2.1,4.0,5.0,6.0
Issue type : kbhowto
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