Last Updated on October 18, 2019 by Admin
CLA – Programming Essentials in C Quizzes Chapter 8 Assessment Exam Answers Full 100%
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA -2 int main(void) { int i = -1; i += ALPHA; printf("%d",i); return 0; }
- the program outputs -1
- the program outputs -2
- the program outputs -3
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA -1+2 int main(void) { int i = -1; i += ALPHA; printf("%d",i); return 0; }
- the program outputs -3
- the program outputs -2
- the program outputs 0
- the program outputs -1
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA -1-2 int main(void) { int i = -1; i = i * ALPHA; printf("%d",i); return 0; }
- the program outputs -1
- the program outputs -2
- the program outputs -3
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA (-1-2) int main(void) { int i = -1; i = -i * ALPHA; printf("%d",i); return 0; }
- the program outputs -2
- the program outputs 0
- the program outputs -3
- the program outputs -1
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA -1 #define BETA - ALPHA int main(void) { int i = ALPHA + BETA * ALPHA * BETA; printf("%d",i); return 0; }
- the program outputs -3
- the program outputs -2
- the program outputs -1
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA(x) -x int main(void) { int i = ALPHA(2-1); printf("%d",i); return 0; }
- the program outputs -1
- the program outputs 0
- the program outputs -2
- the program outputs -3
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA(x) 2*-x int main(void) { int i = ALPHA(1-1); printf("%d",i); return 0; }
- the program outputs -2
- the program outputs -3
- the program outputs -1
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA(x) 2*-x int main(void) { int i = ALPHA((1-1)); printf("%d",i); return 0; }
- the program outputs -1
- the program outputs -2
- the program outputs -3
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA(x,y) x##2-y int main(void) { int i = -1; int i2 = -2; printf("%d",ALPHA(i,i2)); return 0; }
- the program outputs -3
- the program outputs -1
- the program outputs -2
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define ALPHA(x,y) x+y int main(void) { int i = -1; int i2 = -2; printf("%d",-ALPHA(i,i2)); return 0; }
- the program outputs 0
- the program outputs -3
- the program outputs -2
- the program outputs -1
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What happens if you try to compile and run this program?
#include <stdio.h> #define A(x) ((x)?-1:0) int main(void) { int i = 2; int i2 = A(i) * i; printf("%d",i2); return 0; }
- the program outputs -1
- the program outputs -2
- the program outputs 0
- the program outputs -3
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What happens if you try to compile and run this program?
#include <stdio.h> #define A(x) ((x)?-1:0) #define B(a) !(a) int main(void) { int i = 2; int i2 = A(B(i)); printf("%d",i2); return 0; }
- the program outputs -1
- the program outputs -3
- the program outputs -2
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define A(x) #x int main(void) { int i = -1; char *s = A(i); i = -(s[0] == 'i'); printf("%d",i); return 0; }
- the program outputs -2
- the program outputs -3
- the program outputs -1
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define X 1 #define Y 2 int main(void) { int i = #if X<<Y > 0 -X #else -Y #endif ; printf("%d",i); return 0; }
- the program outputs 0
- the program outputs -2
- the program outputs -3
- the program outputs -1
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What happens if you try to compile and run this program?
#include <stdio.h> #define X 1 #define Y 2 int main(void) { int i = #if X>>Y > 0 -X #else -Y #endif ; printf("%d",i); return 0; }
- the program outputs -2
- the program outputs 0
- the program outputs -1
- the program outputs -3
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { int X = 1; #define X 1 int Y = X - 2; #undef X printf("%d",Y+X); return 0; }
- the program outputs -2
- the program outputs -3
- the program outputs 0
- the program outputs -1
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { #undef X int X = 1; int Y = X - 2; #define X -2 printf("%d",Y+X); return 0; }
- the program outputs -2
- the program outputs 0
- the program outputs -1
- the program outputs -3
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What happens if you try to compile and run this program?
#include <stdio.h> #define A #define C int main(void) { int i = #ifdef A #ifdef B -1 #else -2 #endif #else -3 #endif ; printf("%d",i); return 0; }
- the program outputs -2
- the program outputs -3
- the program outputs -1
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> #define A #define C int main(void) { int i = #ifdef A #ifdef C -1 #else -2 #endif #else -3 #endif ; printf("%d",i); return 0; }
- the program outputs -1
- the program outputs -3
- the program outputs 0
- the program outputs -2
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What happens if you try to compile and run this program?
#include <stdio.h> #define B #define C int main(void) { int i = #ifdef A #ifdef C -1 #else -2 #endif #else -3 #endif ; printf("%d",i); return 0; }
- the program outputs -1
- the program outputs -2
- the program outputs 0
- the program outputs -3