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** Assignment statement definition **

** Continuous assignment statement **

Subprogram statements are those used to transfer control between program units—the main program, functions, and subroutines. A function call is performed by invoking the name of the function module in an assignment statement, such as... [Pg.121]

Assignment statements—Assign values by numerical or character constants, by expressions (see Table 1-29 for arithmetic and set operators), or by input to variables, array elements, and fields... [Pg.127]

Thus assignment statements can have any number of entries but only one exit, test statements can have any number of entries but two exits, START statements have no entries and one exit and STOP statements have any number of entries but no exits. [Pg.21]

There are various ways of augmenting the program schemes. For example, a test may have more than two exits, an assignment statement might update several variables simultaneously (e.g. (u,v) (t, t2)) or a term might have functional terms as... [Pg.22]

Let us return once again to Example II-3, and test two paths for consistency using our algorithm. As we did before, we only put down the names of the boxes with assignment statements this uniquely determines each path. [Pg.53]

Notice that T(P) is a proper tree. An execution sequence s of length n 2 2 is consistent with exactly one execution sequence s of length n-1 namely, if s consists of statements (k, k2,...,kn), then s is %, k2,...jk. The tree T(P) is finite branching for if s is an execution sequence of length n and s = (k. .., k ) we need only consider the possibilities far statement k. If it is a STOP statement, s is a complete execution sequence and has no consistent extensions so it labels a node with no sons. If kR is an assignment statement, or a farced transfer, there is exactly one statement kn+1 such that (kpk2,..., k, kn+ ) is consistent, namely the unique statement following. If is a conditional transfer (test) then, since our tests are binary,... [Pg.57]

The construction appears in Example III-3, We allow composite functions in assignment statements for convenience. This can be simulated by simple functions, to meet our definition. That is, we now allow a statement such as u - abaa(u) which can be simulated by applying in order u - a(u), u - a(u), u b(u), and finally u a(u). ... [Pg.72]

In this transformation t we lave changed the order of appearance of tests but the application of assignment statements is unchanged. For each execution sequence in P there is one in t(P) in which the order of applications of assignment statements is the same. If P is liberal, obviously t(P) is liberal. Now in t(P) we have a test Q(u) only after either another test Q (u), ... [Pg.80]

The transformation t we saw at the end of the last section, which changes liberal schemes into free schemes, is such a canonical transformation. The corresponding canonical class of schemes is the class of schemes such that tests are applied initially on the input variables and are applied after assignment statements on the program variables involved, and at no other time. This transformation t is clearly recursive and equivalence preserving. The class of free schemes is not a canonical form class, since, as we saw, there are schemes not strongly equivalent to any free scheme. [Pg.86]

The scheme in P in Example IV-2 is not tree-like. (Here we let A,B,C,D,E stand for arbitrary assignment statements or sets of assignment statements, in order to exhibit the structure of the example uncluttered by extraneous formulae.) The direct connection from the node labelled T(x ) following the node labelled D up to the node labelled B is anomalous since the path START A T(x ) D T(x ) does not contain B and the path START A T(x- ) B does not contain the node labelled T(x- ) following D. Similarly the two direct connections at the bottom of the diagram, from test T(x2) to test T(x ) and from test T(x ) to E are anomalous. The other direct connections in the graph are not anomalous. For example, the direct connection from test T(x2> up to the node labelled A is not anomalous since A is an ancestor of every node in the graph except START. [Pg.105]

Any START statement any STOP statement any assignment statement u t where u is a variable and t is an extended functional term. [Pg.133]

If E(B, m) has been defined and m is labeled with an assignment statement... [Pg.139]

If x E is an assignment statement, x is an identifier for a simple variable and E is an expression of a programming language without side effects but possibly containing x, then the Axiom of Assignment is ... [Pg.179]

Notice that this reverses the way we have been treating assignment statements in path verification conditions. Essentially we have been saying that to prove that B holds of the values after assignment x - E if A holds previously, one must... [Pg.179]

LEMMA 6.16 Let P and P be two free lanov schemes such that P contains n distinct assignment statements and P contains m distinct assignment statements. Schemes P and P are not weakly equivalent if and only if there are paths a in P and a in P beginning at START such that... [Pg.215]

If P and P are free and contain at most r tests, a path with at most nm + 1 assignment statements is altogether of length at most r(nm + 2). Thus we need only test paths of length up to r(nm +2) and they are finite in number. Hence ... [Pg.216]

See also in sourсe #XX -- [ Pg.255 ]

** Assignment statement definition **

** Continuous assignment statement **

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