Answer / susie

Answer :

4 1

Explanation:

The boolean expression needs to be evaluated only till the
truth value of the expression is not known. j is not equal
to zero itself means that the expression’s truth value is 1.
Because it is followed by || and true || (anything) => true
where (anything) will not be evaluated. So the remaining
expression is not evaluated and so the value of i remains
the same.

Similarly when && operator is involved in an expression,
when any of the operands become false, the whole
expression’s truth value becomes false and hence the
remaining expression will not be evaluated.

false && (anything) => false where (anything) will
not be evaluated.

What are the following notations of defining functions known as? i. int abc(int a,float b) { /* some code */ } ii. int abc(a,b) int a; float b; { /* some code*/ }

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1 o 1 1 0 1 0 1 0 1 1 0 1 0 1 how to design this function format in c-language ?

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write a c-program to display the time using FOR loop

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main() { char str1[] = {‘s’,’o’,’m’,’e’}; char str2[] = {‘s’,’o’,’m’,’e’,’\0’}; while (strcmp(str1,str2)) printf(“Strings are not equal\n”); }

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main() { unsigned char i=0; for(;i>=0;i++) ; printf("%d\n",i); }

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abcdedcba abc cba ab ba a a

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main() { struct student { char name[30]; struct date dob; }stud; struct date { int day,month,year; }; scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month, &student.dob.year); }

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void main() { static int i=i++, j=j++, k=k++; printf(“i = %d j = %d k = %d”, i, j, k); }

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#define assert(cond) if(!(cond)) \ (fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\ __FILE__,__LINE__), abort()) void main() { int i = 10; if(i==0) assert(i < 100); else printf("This statement becomes else for if in assert macro"); }

1 Answers  

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