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Recall: Common approaches for choosing test cases • Blackbox testing: Choose based on module’s possible inputs and outputs – Do not use code – Often test boundary cases • White-box testing: Uses internal logic to choose tests – Different levels of code coverage – Aka glass box testing, clear box testing • Regression testing: Keep tests that reveal old bugs – Rationale: “Fixed” bugs come back!

Criteria for choosing test cases: Coverage Measures • Degree to which the source code of a program is tested by a test suite • Examples: – Statement coverage – Condition coverage – Path coverage Some examples will clarify, but first…

Control Flow Graphs – Frequently used to calculate coverage Control Flow Graph int foo(int x, int y) { int z = 0; if ((x>0) && (y>0)) { z = x; } return z; } int z = 0; if ((x>0) && (y>0)) { Basic blocks: straight-line pieces of code without any jumps or jump targets return z; true z = x; false Jumps: control branches

Statement Coverage • Set of test cases such that… Each program statement (line or basic block) is executed at least once

Define a test suite that provides statement coverage for this code Control Flow Graph int foo(int x, int y) { int z = 0; if ((x>0) && (y>0)) { z = x; } return z; } input x 1 y 1 expected 1 ✔ int z = 0; if ((x>0) && (y>0)) { true ✔ z = x; ✔ return false z;

Now try this code… Control Flow Graph int result=0; result = x*2; x < 69 T F x = 6969; x % 2 == 0 T F x > 6969 F public int bar(int x) { int result=0; if (x < 69) { if (x % 2 == 0) { result = x/2; } else { result = x*2; } } else if (x > 6969) { x = 69; } else { x = 6969; } return result; } result = x/2; T x = 69; return result;

Now try this code… ✔int result=0; < 69 T > 6969 F = 6969; ✔ result ✔x 6970 70 69 6969 T ✔x = x*2; F % 2 == 0 ✔result F ✔x expected 2 1 Control Flow Graph ✔x ✔x Input (x) 1 2 T = x/2; = 69; ✔ return result;

Condition Coverage • Set of test cases such that… Each boolean expression (in control structures) evaluates to true at least once and to false at least once

Define a test suite that provides condition coverage for this code Control Flow Graph int foo(int x, int y) { int z = 0; if ((x>0) && (y>0)) { z = x; } return z; } input x 1 0 y 1 0 expected 1 0 int z = 0; if ((x>0) && (y>0)) { ✔true z = x; return z; ✔false

Now try this code… Control Flow Graph int result=0; result = x*2; x < 69 T F x = 6969; x % 2 == 0 T F x > 6969 F public int bar(int x) { int result=0; if (x < 69) { if (x % 2 == 0) { result = x/2; } else { result = x*2; } } else if (x > 6969) { x = 69; } else { x = 6969; } return result; } result = x/2; T x = 69; return result;

Now try this code… Control Flow Graph Input (x) 1 2 6970 70 int result=0; result = x*2; x < 69 ✔ x = 6969; x % 2 == 0 ✔ ✔ F F T F x > 6969 ✔ ✔ T result = x/2; ✔x T = 69; return result; expected 2 1 69 6969

Path Coverage • Set of test cases such that… Each possible path through a program’s control flow graph is taken at least once

Define a test suite that provides path coverage for this code Control Flow Graph int foo(int x, int y) { int z = 0; if ((x>0) && (y>0)) { z = x; } return z; } input x 1 0 y 1 0 expected 1 0 int z = 0; if ((x>0) && (y>0)) { true (a) z = x; (b) return z; Paths: ✔ • a , b ✔ • c (c) false

Now try this code… Control Flow Graph int result=0; result = x*2; x < 69 T F x = 6969; x % 2 == 0 T F x > 6969 F public int bar(int x) { int result=0; if (x < 69) { if (x % 2 == 0) { result = x/2; } else { result = x*2; } } else if (x > 6969) { x = 69; } else { x = 6969; } return result; } result = x/2; T x = 69; return result;

Now try this code… Control Flow Graph Input (x) expected 1 2 6970 70 2 1 69 6969 int result=0; result = x*2; (a) x < 69 (b) F x > 6969 (c) F x = 6969; T (d) (f) F x % 2 == 0 (h) (g) T result = x/2; T x = 69; (e) (k) (i) (j) Paths: ✔ • a , d , f , h ✔ • a , d , g , i ✔ • a , b , e , j ✔ • a , b , c , k return result;

BONUS QUESTION

Given this code 1. Draw a control flow graph 2. Define test suites that provide: • Statement coverage • Condition coverage • Path coverage int my. F(int x, int y) { while (x > 10) { x = x – 10; if (x == 10) { break; } } if (y < 20 && x%2 == 0) { y = y + 20; } else { y = y – 20; } return 2*x + y; }

int my. F(int x, int y) { while (x > 10) { x = x – 10; if (x == 10) { break; } } if (y < 20 && x%2 == 0) { y = y + 20; } else { y = y – 20; } return 2*x + y; } while (x > 10) T F x = x – 10; if (x == 10) T break; if (y < 20 && x%2 == 0) F y = y – 20; Here’s the CFG F T y = y + 20; return 2*x + y;