Last Updated on October 18, 2019 by Admin
CLA – Programming Essentials in C Quizzes Chapter 3 Assessment Exam Answers Full 100%
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { int i, j, k; i = -1; j = 1; if(i) j--; if(j) i++; k = i * j; printf("%d",k); return 0; }
- the program outputs2
- the program outputs -1
- 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> int main(void) { int i, j, k; i = 0; j = 0; if(j) j--; else i++; if(i) i--; else j++; k = i + j; printf("%d",k); return 0; }
- the program outputs 2
- the program outputs -1
- 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) { int i, j, k; i = 2; j = 3; if(j) j--; else if(i) i++; else j++; if(j) i--; else if(j) j++; else j = 0; k = i + j; printf("%d",k); 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> int main(void) { double x = -.1; int i = x; printf("%d",i); return 0; }
- the program outputs -0.100000
- the program outputs -1
- the program outputs 0.100000
- the program outputs 0
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { float x,y; int i,j; x = 1.5; y = 2.0; i = 2; j = 3; x = x * y + i / j; printf("%f",x); return 0; }
- the program outputs 3.000000
- the program outputs 2.000000
- the program outputs 1.000000
- the program outputs 0.000000
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { float x,y; int i,j; x = 1.5; y = 2.0; i = 2; j = 4; x = x * y + (float)i / j; printf("%f",x); return 0; }
- the program outputs 2.000000
- the program outputs 3.000000
- the program outputs 4.000000
- the program outputs 3.500000
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { int i; i = 1; while(i < 16) i *= 2; printf("%d",i); return 0; }
- the program outputs 32
- the program outputs 4
- the program outputs 16
- the program outputs 8
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { int i, j; i = 1; j = 1; while(i < 16) { i += 4; j++; } printf("%d",j); return 0; }
- the program outputs 4
- the program outputs 7
- the program outputs 5
- the program outputs 6
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { int i = 7, j = i - i; while(i) { i /= 2; j++; } printf("%d",j); 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> int main(void) { int i = 7, j = i - i; while(!i) { i /= 2; j++; } printf("%d",j); 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> int main(void) { int i, j = 1; for(i = 11; i > 0; i /= 3) j++; printf("%d",j); return 0; }
- the program outputs 3
- the program outputs 5
- the program outputs 4
- the program outputs 2
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What happens if you try to compile and run this program?
#include <stdio.h> int main(void) { int i, j = 0; for(i = 0; !i ; i++) j++; printf("%d",j); 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> int main(void) { int i = 1, j = -2; for(;;) { i *= 3; j++; if(i > 30) break; } printf("%d",j); return 0; }
- the program outputs 3
- the program outputs 1
- 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> int main(void) { int i = 1, j = -2, k; k = (i >= 0) && (j >= 00) || (i <= 0) && (j <= 0); printf("%d",k); 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> int main(void) { int i = 1, j = -2, k; k = (i >= 0) || (j >= 00) && (i <= 0) || (j <= 0); printf("%d",k); 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> int main(void) { int i = 1, j = -2, k; k = !(i >= 0) || !(j >= 00) && !(i <= 0) || !(j <= 0); printf("%d",k); return 0; }
- the program outputs 0
- the program outputs 1
- 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> int main(void) { int i = 1, j = 0, k; k = i & j; k |= !!k; printf("%d",k); 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> int main(void) { int i = 1, j = 0, k; k = !i | j; k = !k; printf("%d",k); 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 i = 1, j = 0, k; k = (i ^ j) + (!i ^ j) + (i ^ !j) + (!i ^ !j); printf("%d",k); 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> int main(void) { int i = 0, j = 1, k; k = i << j + j << i; printf("%d",k); return 0; }
- the program outputs 0
- the program outputs 2
- the program outputs 3
- the program outputs 1