CLA – Programming Essentials in C Quizzes Final Test Online

What is the value of the following integer literal?

    012

What is the value of the following integer literal?

    0x12

What is the value of the X variable at the end of the following snippet?

    int X = 1;

    X = X + 2 * X;
    X = X / 2 * X;
    X = X + 2 + X;

What is the value of the X variable at the end of the following snippet?

    int X = 1;

    X = X * X + 2;
    X = X / X * 2;
    X = X + 2 + X;

What is the value of the X variable at the end of the following snippet?

    int X = 1, Y = 2, Z;
    
    Z = X / Y * --X * Y++;

What is the value of the X variable at the end of the following snippet?

    int X;

    X = 'b' - 'a' * ('\' / '\');

What happens if you try to compile and run this program?

    #include <stdio.h>
    int main(void) {
        int a = -1, b = 1;
        float i = 2.0, j = -2.0;
        printf("%d\n", (a > b) + (b > a) + (i > j) + (j > i) + ('z' > 'a'));
        return 0;
    }

What happens if you try to compile and run this program?

    #include <stdio.h>
    int main(void) {
        int i = 0;
        if(i = 1)
        	i = 2;
        else
        	i = 3;
        printf("%d\n",i);
        return 0;
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        float x = 3.0, y = 2.0; 
        int i = 1, j = 2; 
        x = (int)x / y + (float)i / j; 
        printf("%f",x); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i = 16, j = 8;
        do {
        	i /= 2;
        	j -= i / 2;
        } while(j > 0);
        printf("%d",i + j); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i = 16, j = 6;
        while(j > 0) {
        	i /= 2;
        	j -= i / 2;
        } 
        printf("%d",i + j); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i = 2, j;
        for(j = 0; j < 0; j -= i) 
        	i /= 2;
        printf("%d",i + j); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i = 5, j = 4;
        for(i--; i--; i--) 
        	j--;
        printf("%d",i + j); 
        return 0; 
    }

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; 
    }

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) + (j << i) + (i << i) + (j << j);
        k >>= i;
        printf("%d", k); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i = 1, j = i + 2 * i;
        switch(j - i) {
        	case  1: j++;
        	case  2: j--;
        	case  0: j++; break;
        	default: j = 0;
        }
        printf("%d", ++j); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i = 1, j = i + 2 * i;
        switch(j) {
        	default: j = 0;
        	case  1: j++; break;
        	case  2: j--;
        	case  0: j++; break;
        }
        printf("%d", ++j); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i, t[4];
        for(i = 0; i < 3; i++) {
        	t[i] = i;
        	t[i + 1] = 2 * t[i];
        }
        printf("%d\n", t[3]);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i, t[4];
        for(i = 3; i; i--) {
        	t[i] = i - 1;
        	t[t[i]] = t[i];
        }
        printf("%d\n", t[0]);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int i, s = 0, t[] = {0, 1, 2, 4, 8, 16};
        for(i = 2; t[i] < 8; i *= 2) 
        	s += t[i];
        printf("%d\n", s);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        char t[] = { 'a', 'z', 'B', 'Z', '0' };
        printf("%d\n", t[t[1] - t[0] - t[3] + t[2] + 3] - t[4]);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        char a = 'A', *b = &a, **c = &b;
        **c = a + (a == *b);
        printf("%c", a);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int t[4][4];
        printf("%d\n",sizeof(t) / sizeof(t[0]) / sizeof(t[0][0]));
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        int t[4] = { 0, -1, -2, -3 }, *p = t + 3;
        printf("%d\n", p[*p] - t[2]);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        char *p = "\0\4\1\3\2";
        printf("%d\n", p[p[1]] + *(p + 1) + p[4]);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #include <string.h>
    int main(void) { 
        char tt[20] = "0123456789";
        strcat(tt + 2, "987");
        printf("%d\n", strlen(tt) - tt[5] + '0');
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #include <string.h>
    int main(void) { 
        char tt[20] = "0123456789";
        strcpy(tt, tt + 2);
        printf("%d\n", strlen(tt) - tt[9] + '5');
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #include <stdlib.h>
    int main(void) { 
        int *t = (int *) malloc(sizeof(int) + sizeof(int));
        t++;
        *t = 8;
        t[-1] = *t / 2;
        t--;
        t[1] = *t / 2;
        printf("%d\n",*t);
        free(t);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #include <string.h>
    struct S {
        char S[4];
    };
    int main(void) { 
        struct S S = { 'a', 'b' };
        printf("%d", sizeof(S.S) - strlen(S.S) + S.S[3]); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #include <string.h>
    #include <stdlib.h>
    struct S {
        char *S;
    };

    int main(void) { 
        struct S *S = (struct S *) malloc(sizeof(struct S));
        S -> S = "abc";
        printf("%d", strlen(S -> S + 2) + S -> S[3]); 
        free(S);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    struct S {
        int Var;
        struct S *Str;
    };
    int main(void) { 
        struct S S[] = { { 8, NULL }, { 4, &S[0] }, { 2, &S[1] } };
        printf("%d", S[2].Str->Str->Var);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int fun(int *t) {
        return *(++t);
    }
    int main(void) { 
        int arr[] = { 8, 4, 2, 1 };
        printf("%d\n", fun(arr + 2));
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int fun(int t) {
        return ++t;
    }
    int main(void) { 
        int arr[] = { 8, 4, 2, 1 };
        printf("%d\n", fun(arr[3]) + arr[2]);
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int f(int v) { 
        	v = 2 * v;
        return v * v;
    } 
    int main(void) { 
        int i = 2; 
        f(i); 
        printf("%d",i); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    char *f(char *p) { 
        return p++;
    } 
    char *g(char *p) {
        return p += 2;
    }
    int main(void) { 
        char *s = "ABCDEFGHIJ";
        char  p = *f(g(f(s + 6)));
        printf("%d",p - 'A'); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    struct S {
        int S[2];
    };
    void f(struct S S) { 
        S.S[0] = S.S[1] + 4;
    } 
    int main(void) { 
        struct S S = { { 4, 8 } };
        f(S);
        printf("%d",S.S[1] / S.S[0]); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    struct S {
        int S[2];
    };
    void f(struct S *S) { 
        S->S[1] = S->S[0] + 2;
    } 
    int main(void) { 
        struct S S = { { 4, 8 } }, *P = &S;
        f(P);
        printf("%d",S.S[1] / S.S[0]); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int f(int t[][2]) { 
        return t[0][0] + t[0][1]; 
    } 
    int main(void) { 
        int i,t[2][2] = { {0,4},{4,2} }; 
        i = f(t); 
        printf("%d",i); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #include <string.h>
    char *f(int p, char *s) { 
        s[p + 1] = '\0';
        return s + 1;
    } 
    int main(void) { 
        char s[] = "ABCDEF";
        int i = strlen(f(1,s + 2));
        printf("%d\n",i); 
        return 0; 
    }

What happens if you try to compile and run this program?

    #include <stdio.h> 
    int main(void) { 
        char s[20]; 
        FILE *f = fopen("data","w"); 
        int i = fputs("1248",f); 
        fclose(f); 
        f = fopen("data","r"); 
        fgets(s + 2,4,f); 
        putchar(s[4]); 
        fclose(f); 
        return 0;

What happens if you try to compile and run this program?

    #include <stdio.h> 
    #define  ABC    10
    #define  XYZ 	ABC - 1
    int main(void) { 
        int i = 19;
        i = i - XYZ;
        printf("%d\n", i);
        return 0; 
    }

What is the meaning of the following declaration?

    void (*f)(int);