Understanding of software test logic coverage

determine 、 Conditions 、 Use cases

for(int i=2;i<=n && k==1;i++){
    if(m%i==0){
        break;
    }
}

The above code means when $k=1$ when , find $[2,n]$ Within the interval $m$ The first factor of , Take this as an example

Judgment and condition

determine , Such as if、for、while And so on

Conditions （ Logical conditions ）, And one Logical judgment statement It corresponds to a Basic conditions

A number of Basic conditions It can be done through 、 or 、 Unequal logical connectives form Compound condition

Be careful , The question is to meet the needs of coverage Logical conditions when , According to the meaning of the question, you can judge by yourself Basic conditions still Compound condition

• As in the above code Logical judgment sub statement yes i<=n and k==1 and m%i==0

• If the above code is satisfied 100% Determine what is required for coverage Logical conditions yes i<=n And k=1 and i>n or k≠1 and m%i=0 and m%i≠0

• If the above code is satisfied 100% Required for conditional coverage Logical conditions yes i<=n and i>n and k=1 and k≠1 and m%i=0 and m%i≠0

in other words , Yes $n$ individual Basic conditions ,$n$ individual Basic conditions constitute $m$ individual Compound condition

• To meet the 100% Decision coverage rule most Yes $2m$ Logical conditions
• To meet the 100% Conditional coverage is most Yes $2n$ Logical conditions

The test case

The test case It refers to the specific value of a group of variables , One The test case It may be possible to satisfy multiple Logical conditions

What we need to see clearly is 100% Cover the required Logical conditions Or at least The test case , One The test case It may be possible to satisfy multiple Logical conditions

If the above code is satisfied 100% One set of test cases covered by the condition is

• {n=2、m=3、k=1} ： Satisfy i<=n And k=1 and m%i=0 and m%i≠0
• {n=0、m=3、k=1} ： Satisfy i>n or k≠1

If the above code is satisfied 100% One set of test cases covered by the condition is

• {n=2、m=3、k=1}： Satisfy i<=n and k=1 and m%i=0 and m%i≠0
• {n=0、m=3、k=0}： Satisfy i>n and k≠1

The number of required logical conditions is the same as Minimum test cases Quantity is not directly related to …… Because a test case may satisfy several logical conditions

Real example

void cal( int n ) { 
    int g, s, b, q; 
    if ( ( n > 1000 ) && ( n < 2000 ) ) { 
        g = n % 10; 
        s = n % 100 / 10; 
        b = n / 100 % 10; 
        q = n / 1000; 

        if( ( q + g ) == ( s + b ) ) { 
            printf("%-5d", n); 
        } 
    } 

   return; 
} 

Logical coverage

The logic coverage part of this article is simple , You can check this article for details

《 White box test case design method 》

https://blog.csdn.net/tianlin151/article/details/115665358

Logical coverage includes the following ：

• Statement override （Statement Coverage, SC）
• Determine coverage （Decision Coverage, DC）
• Conditional coverage （Condition Coverage,CC）
• Judge whether the condition covers （Condition/ Decision Coverage, CDC）
• Conditional combination covering （Multiple Condition Coverage, MCC）
• Path coverage （Path Coverage, PC）

void f(double x, double y, double z){
    double num1, num2;// Sentence block 1
    if(x>=0 && y!=0){
        num1=sqrt(x)/y;// Sentence block 2
        printf("num1 is %f\n",num1);// Sentence block 2
    }
    else{
        printf("num1 is invalid\n");// Sentence block 3
    }
    if(z>=0){
        num2=z;// Sentence block 4
        printf("num2 is %f\n",num2);// Sentence block 4
    }
    else{
        num2=-z;// Sentence block 5
        printf("num2 is %f\n",num2);// Sentence block 5
    }
    printf("end of calculation\n");// Sentence block 6
    printf("goodbye\n");// Sentence block 6
}

The above code means to calculate in sequence $\frac{\sqrt{x}}{y}$ and $|z|$, The flow chart is as follows , The following text takes this as an example

• if(x>=0 && y!=0) determine , Write it down as P1
• if(z>=0) determine , Write it down as P2
• x>=0 Conditions , Write it down as C1
• y!=0 Conditions , Write it down as C2
• z>=0 Conditions , Write it down as C3

Statement override

Design test cases , Each statement in the program is executed at least once .

For the example above 100% Statement coverage requires statement blocks 1、 Sentence block 2、 Sentence block 3、 Sentence block 4、 Sentence block 5、 Sentence block 6 Are executed at least once

Determine coverage （ Branch coverage ）

Design test cases , Make every judgment in the program ” really “ and ” false “ Are executed at least once . namely ： Each branch in the program executes at least once .

For example 100% To determine the coverage is to P1、P2 Each is true and false at least once

Accessible 100% Determine the satisfaction of one set of test cases covered ：

• The test case 1： At the same time satisfy P1 really 、P2 really
• The test case 2： At the same time satisfy P1 false 、P2 false

Conditional coverage

Design test cases , So that each condition in the decision takes the true value at least once , Once take a false value .

For example 100% Conditional coverage is the need to C1、C2、C3 Each is true and false at least once

Accessible 100% The satisfaction of one set of test cases covered by the condition ：

• The test case 1： At the same time satisfy C1 really 、C2 really 、C3 really
• The test case 2： At the same time satisfy C1 false 、C2 false 、C3 false

determine - Conditional coverage

Design test cases , Make the judgment result of each judgment in the tested program （ True and false ） At least once , meanwhile , Possible values of each logical condition （ True and false ） And be satisfied at least once . That is to satisfy at the same time 100% Determine coverage and 100% Criteria for conditional coverage .

For example 100% determine - Conditional coverage is the need to P1、P2、C1、C2、C3 Each is true and false at least once

Accessible 100% determine - The satisfaction of one set of test cases covered by the condition ：

• The test case 1： At the same time satisfy C1 really 、C2 really 、P1 really 、C3 really 、P2 really
• The test case 2： At the same time satisfy C1 false 、C2 false 、P1 false 、C3 false 、P2 false

Conditional combination covering

Design test cases , Make all possible combinations of condition results in each decision in the tested program execute at least once .

in other words , Yes $n$ individual Basic conditions , Every Basic conditions There is a true and a false 2 In this case , When combined with each other, there will be $2^n$ Test cases

Accessible 100% The satisfaction of one set of test cases covered by the condition combination ：

• The test case 1：C1 really 、C2 really 、C3 really
• The test case 2：C1 really 、C2 really 、C3 false
• The test case 3：C1 really 、C2 false 、C3 really
• The test case 4：C1 really 、C2 false 、C3 false
• The test case 5：C1 false 、C2 really 、C3 really
• The test case 6：C1 false 、C2 really 、C3 false
• The test case 7：C1 false 、C2 false 、C3 really
• The test case 8：C1 false 、C2 false 、C3 false

Path coverage

Design test cases , Covering all possible paths in the program .

For example, the path is required a、b、c、d、e、f、g、h、i、j、k Walk at least once