INFO: Difference Between Arrays and Pointers in CLast reviewed: September 2, 1997Article ID: Q44463 |
The information in this article applies to:
SUMMARYThe 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 INFORMATIONThe 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.
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Additional query words: 8.00
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