This is the problems for unix, Please make sure the answer is correct, thank u!

Question

This is the problems for unix, Please make sure the answer is correct, thank u!

(a) (4 marks) In which phase of compilation is the following line of a C program processed: #include and how it affects further compilation? (b) (4 marks) Describe what happens when the following scanf statement: scanf ("%d%d%d", &i;, &j;, &k;); reads the folowing input: 1.2.3 What is the return value of the scanf function? ata Iine is provided in input as follows: 2017/10/06 23.45 Write a scanf statement to read the numbers above in the integer variables (y, m, and d) and one floating point variable (x), so the assigned values after scanf are as follows: y-2017 m-10 d-06 x-23.45 (d) (4 marks) Assume that four variables are given as in part (c). Write a printf statement with the integer variables y, m, d, and the floating point variable x, to produce an output as analysed in part (c), e.g., somethink like 2017/10/06 23.45 Have in mind that the variables m and d must have two digits with possibly a leading 0, and the variable x must be printed with the precisio of two digits.

Explanation / Answer

1)

In compilation during the stage of preprocessor at starting of program, #include is replaced by it's text/content.

2)

scanf reads the values 1,2,3 and stores in i,j,k respectively.

And the return of scanf will be number of items read. i.e. 3 in this case.

3)

int y,m,d;

float x;

// %d to read int and %f to read float

scanf("%d%d%d%f",&y,&m,&d,&x);

4)

// %02d adds a leading zero if length less than 2. %.2f prints only two decimal places
printf("%d/%02d/%02d %.2f ",y,m,d,x);

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