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...main() { int a[4]; int b[3]={1,2,3}; int i,j; for(i=0;i<4;i...
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Given this memory layout, &a[2] - &b[3] is +3 because a[2] is three elements later in memory than b[3]. Then &b[3] - &a[2] is -3 because b[3] is three elements earlier in memory than a[2]. (Note that simple pointer arithmetic like this that does not conform to the C standard can break because the compiler may perform various optimizations or ...
1. What will be the output of following program? #include int main() { int a = 5, *b, c; b = &a; printf("%d", a * *b * a + *b); return (0); } Options: 1. 130 2. 103 3. 100 4. 310 The answer is the option(1). Explanation: Here the expression a**b*a + *b uses pointer in C
Answer : 7, 224, 0 Description : As x = 7 so first %d gives 7, second %d will take value of x after left shifting it five times, and shifting is done after converting the values to binary, binary value of 7 (000111) will be left shifted twice to make it binary 224(11100000), so x<<5 is 224 and as left shifting does not effect the original value of x its still 5 so third %d will also show 0.
Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3. Step 2: a = CUBE(b++); becomes => a = b++ * b++ * b++; => a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement. => a = 27; Here, 27 is store in the variable a. By the way, the value of ...
C Programming questions and answers section on "Control Instructions Find Output of Program" for placement interviews and competitive exams: Fully solved C Programming problems with detailed answer descriptions and explanations are given for the "Control Instructions Find Output of Program" section - Page 4.
static int a[] = {10, 20, 30, 40, 50}; static int *p[] = {a, a+3, a+4, a+1, a+2}; int **ptr = p; ptr++; $\text{ptr-p} = \frac{\text{address of ptr} - \text{address of ...
SYNTAX of 3-D array : data_type arr[Block_size][row_size][column_size];we can also solve this question by keeping in mind the above syntax :)
1. What will be the output of following program? #include int main() { int a = 5, *b, c; b = &a; printf("%d", a * *b * a + *b); return (0); } Options: 1. 130 2. 103 3. 100 4. 310 The answer is the option(1). Explanation: Here the expression a**b*a + *b uses pointer in C
void main() {int a[3]; a[0] = 4; a[1] = 5; a[2] = 6;} a[0] a a[1] a[2] 4 5 6 activation record main() • The type of a is (int *). • As a pointer, a points to the first memory location in the array. • Stay tuned for details... 4. Arrays in C • No bounds checking. Make sure that you only ac-
1. What will be the output of following program? #include int main() { int a = 5, *b, c; b = &a; printf("%d", a * *b * a + *b); return (0); } Options: 1. 130 2. 103 3. 100 4. 310 The answer is the option(1). Explanation: Here the expression a**b*a + *b uses pointer in C