Use-define chain

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A Use-Definition Chain (UD Chain) is a data structure that consists of a use, U, of a variable, and all the definitions, D, of that variable that can reach that use without any other intervening definitions. A UD Chain generally means the assignment of some value to a variable.

A counterpart of a UD Chain is a Definition-Use Chain (DU Chain), which consists of a definition, D, of a variable and all the uses, U, reachable from that definition without any other intervening definitions.

Both UD and DU chains are created by using a form of static code analysis known as data flow analysis. Knowing the use-def and def-use chains for a program or subprogram is a prerequisite for many compiler optimizations, including constant propagation and common subexpression elimination.


Making the use-define or define-use chains is a step in liveness analysis, so that logical representations of all the variables can be identified and tracked through the code.

Consider the following snippet of code:

 int x = 0;    /* A */
 x = x + y;    /* B */
 /* 1, some uses of x */
 x = 35;       /* C */
 /* 2, some more uses of x */

Notice that x is assigned a value at three points (marked A, B, and C). However, at the point marked "1", the use-def chain for x should indicate that its current value must have come from line B (and its value at line B must have come from line A). Contrariwise, at the point marked "2", the use-def chain for x indicates that its current value must have come from line C. Since the value of the x in block 2 does not depend on any definitions in block 1 or earlier, x might as well be a different variable there; practically speaking, it is a different variable — call it x2.

 int x = 0;    /* A */
 x = x + y;    /* B */
 /* 1, some uses of x */
 int x2 = 35;  /* C */
 /* 2, some uses of x2 */

The process of splitting x into two separate variables is called live range splitting. See also static single assignment form.


The list of statements determines a strong order among statements.

  • Statements are labeled using the following conventions: , where i is an integer in ; and n is the number of statements in the basic block
  • Variables are identified in italic (e.g., v,u and t)
  • Every variable is assumed to have a definition in the context or scope. (In static single assignment form, use-define chains are explicit because each chain contains a single element.)

For a variable, such as v, its declaration is identified as V (italic capital letter), and for short, its declaration is identified as . In general, a declaration of a variable can be in an outer scope (e.g., a global variable).

Definition of a Variable[edit]

When a variable, v, is on the LHS of an assignment statement, such as , then is a definition of v. Every variable (v) has at least one definition by its declaration (V) (or initialization).

Use of a Variable[edit]

If variable, v, is on the RHS of statement , there is a statement, with i < j and , that it is a definition of v and it has a use at (or, in short, when a variable, v, is on the RHS of a statement , then v has a use at statement ).


Consider the sequential execution of the list of statements, , and what can now be observed as the computation at statement, j:

  • A definition at statement with i < j is alive at j, if it has a use at a statement with kj. The set of alive definitions at statement i is denoted as and the number of alive definitions as . ( is a simple but powerful concept: theoretical and practical results in space complexity theory, access complexity(I/O complexity), register allocation and cache locality exploitation are based on .)
  • A definition at statement kills all previous definitions ( with k < i) for the same variables.

Execution example for def-use-chain[edit]

This example is based on a Java algorithm for finding the gcd. (It is not important to understand what this function does.)

 2 * @param(a, b) The values used to calculate the divisor.
 3 * @return The greatest common divisor of a and b.
 4 */
 5int gcd(int a, int b) { 
 6    int c = a;
 7    int d = b; 
 8    if (c == 0)
 9        return d;
10    while (d != 0) { 
11        if (c > d)
12            c = c - d;
13        else
14            d = d - c;
15    } 
16    return c; 

To find out all def-use-chains for variable d, do the following steps:

  1. Search for the first time the variable is defined (write access).
    In this case it is "d=b" (l.7)
  2. Search for the first time the variable is read.
    In this case it is "return d"
  3. Write down this information in the following style: [name of the variable you are creating a def-use-chain for, the concrete write access, the concrete read access]
    In this case it is: [d, d=b, return d]

Repeat these steps in the following style: combine each write access with each read access (but NOT the other way round).

The result should be:

1 [d, d=b, return d] 
2 [d, d=b, while(d!=0)] 
3 [d, d=b, if(c>d)] 
4 [d, d=b, c=c-d] 
5 [d, d=b, d=d-c]
6 [d, d=d-c, while(d!=0)] 
7 [d, d=d-c, if(c>d)] 
8 [d, d=d-c, c=c-d] 
9 [d, d=d-c, d=d-c]

You have to take care, if the variable is changed by the time.

For example: From line 7 down to line 13 in the source code, d is not redefined / changed. At line 14, d could be redefined, this is, why you have to recombine this write access on d with all possible read access, which could be reached. In this case, only the code beyond line 10 is relevant. Line 7 for example cannot be reached again. For your understanding, you can imagine 2 different variables d:

1 [d1, d1=b, return d1] 
2 [d1, d1=b, while(d1!=0)] 
3 [d1, d1=b, if(c>d1)] 
4 [d1, d1=b, c=c-d1] 
5 [d1, d1=b, d1=d1-c]
6 [d2, d2=d2-c, while(d2!=0)] 
7 [d2, d2=d2-c, if(c>d2)] 
8 [d2, d2=d2-c, c=c-d2] 
9 [d2, d2=d2-c, d2=d2-c]

As result you could get something like this. The variable d1 would be replaced by b

 2 * @param(a, b) The values used to calculate the divisor.
 3 * @return The greatest common divisor of a and b.
 4 **/
 5int gcd(int a, int b) {
 6    int c = a;
 7    int d; 
 8    if (c == 0)
 9        return b;
10    if (b != 0) {
11        if (c > b) {
12            c = c - b;
13            d = b;
14        }
15        else
16            d = b - c;
17        while (d != 0) { 
18            if (c > d)
19                c = c - d;
20            else
21                d = d - c;
22        }
23    } 
24    return c; 

Method of building a use-def (or ud) chain[edit]

  1. Set definitions in statement
  2. For each i in , find live definitions that have use in statement
  3. Make a link among definitions and uses
  4. Set the statement , as definition statement
  5. Kill previous definitions

With this algorithm, two things are accomplished:

  1. A directed acyclic graph (DAG) is created on the variable uses and definitions. The DAG specifies a data dependency among assignment statements, as well as a partial order (therefore parallelism among statements).
  2. When statement is reached, there is a list of live variable assignments. If only one assignment is live, for example, constant propagation might be used.