01
The ASCII value of the digit character ‘0’ is 48. Given the following program:
main()
{
char a=’1′,b=’2′;
printf(“%c,”,b++);
printf(“%d\n”,b-a);
}
The output of the program is:
A) 3,2
B) 25,2
C) 2,2
D) 2,25
Answer: C
Explanation:
printf(“%c, b++): The initial value of b is ‘2’ (ASCII value 50); b++ is a post-increment, so it first outputs the current value of b ‘2’, then b becomes ‘3’ (ASCII value 51), resulting in the output 2,.
printf(“%d\n”, b – a): At this point, b is ‘3’ (ASCII value 51) and a is ‘1’ (ASCII value 49); b – a calculates the difference in ASCII values: 51 – 49 = 2, resulting in the output 2.
The combined output of the two printf statements is 2,2
02
Read the following program segment:
int a, b, c;
a=10; b=50; c=a;
if (a>b) a=b, b=c; c=a;
printf(“a=%d b=%d c=%d\n”, a, b, c);
Then the output of the program is ( )
A) a=10 b=50 c=10
B) a=10 b=50 c=30
C) a=10 b=30 c=10
D) a=50 b=30 c=50
Answer: A
Explanation:
Variable initialization:
a = 10;
b = 50;
c = a; (at this point c = 10)
if (a > b) check:
a is not greater than b, so the code inside the if statement a = b, b = c; will not execute.
c = a; executes; regardless of whether the if condition is true, c = a; will execute, so c is reassigned to 10.
Final variable values:
a remains 10
b remains 50
c is updated to 10
03
The output of the following program segment is
char a[ ]= “language”, *p;
p=a;
while (*p !=’u’) { printf(“%c”,*p-32); p++; }
A) LANGUAGE
B) language
C) LANG
D) langUAGE
Answer: C
Explanation:
while (*p != ‘u’) loop:
Loop condition: continue executing while the current *p is not equal to ‘u’.
In “language”, ‘u’ is the 5th character (counting from 0, it’s a[4]), so the loop will process the first 4 characters ‘l’, ‘a’, ‘n’, ‘g’.
*p – 32: converts lowercase letters to uppercase letters (in ASCII, lowercase letters are 32 greater than uppercase letters).
‘l’ – 32 = ‘L’
‘a’ – 32 = ‘A’
‘n’ – 32 = ‘N’
‘g’ – 32 = ‘G’
p++: moves the pointer to the next character; when p points to ‘u’, the loop ends, and LANG is output, and the program only outputs LANG, not processing the subsequent characters.
04 (Common Mistake)
Given int a; float b; when executing scanf(“%2d%f”,&a,&b); if the input from the keyboard is 876 543.0<enter>, the values of a and b are:
A) 876 and 543.000000
B) 87 and 6.000000
C) 87 and 543.000000
D) 76 and 543.000000
Answer: B
Explanation:
%2d format specifier: %2d means read at most 2 digits of an integer, so the first two digits of 876, 87, are assigned to a, and the remaining 6 will be processed by the subsequent %f, so a = 87.
%f format specifier: %f will read the remaining input part, i.e., 6 543.0, since there is a space after 6, %f will read 6 and stop (as space is the default delimiter), so b = 6.000000; the remaining 543.0 will be left in the input buffer but will not be assigned to any variable.
05
Let the variables be defined as int A=5, B=6, the value of the expression (++A==B–)?++A:–B is
A) 5
B) 6
C) 7
D) 8
Answer: C
Explanation:
Initial values A = 5, B = 6
Evaluating the expression (++A == B–)
++A: first increment A by 1, A becomes 6, then returns 6.
B–: first returns the current value of B 6, then B decrements to 5.
Comparing 6 == 6, the result is true (1).
Execution of the ternary operator ? :
Since (++A == B–) is true, execute ++A. ++A: at this point A is already 6, incrementing it by 1 makes it 7, returning 7; –B will not execute; the final result: the value of the entire expression is 7.
Conditional operator:
Conditional expression?expression1:expression2
The ternary operator is the only three-way operator in C language, its syntax rules are to first evaluate the conditional expression (which must return true or false), if the condition is true, execute expression1 and return its value; if the condition is false, execute expression2 and return its value.
06
If the following program exists:
int a[12]={1,2,3,4,5,6,7,8,9,10,11,12};
char c=’a’,d,g;
Then the expression that evaluates to 4 is
A) a[g-c]
B) a[4]
C) a[‘d’-‘c’]
D) a[‘d’-c]
Answer: D
Explanation:
For option A: g is uninitialized, its value is uncertain, so the result of a[g – c] is also uncertain.
For option B: array indexing starts from 0, a[4] is the 5th element, which is 5.
For option C: the ASCII value of ‘d’ is 100, and ‘c’ is 99; ‘d’ – ‘c’ = 100 – 99 = 1; a[1] is the 2nd element, which is 2.
For option D: c = ‘a’, its ASCII value is 97; ‘d’ – c = 100 – 97 = 3; a[3] is the 4th element, which is 4.
07
Read the following program:
int fun(int u,int v);
main ()
{
int a=24,b=16,c;
c=fun(a,b);
printf(“%d\n”,c);
}
int fun(int u,int v)
{ int w;
while(v)
{ w=u%v; u=v; v=w }
return u;
}
The output is
A) 6
B) 7
C) 8
D) 9
Answer: C
Explanation:
Loop iterations u v w u’ v’
1 24 16 8 16 8
2 16 8 0 8 0
End 8 0 Loop terminates
08
Read the following program:
main()
{ char a,b,c,d;
scanf(“%c,%c,%d,%d”,&a,&b,&c,&d);
printf(“%c,%c,%c,%c\n”,a,b,c,d);
}
If the input from the keyboard is: 6,5,65,66<enter>, the output is
A) 6,5,A,B
B) 6,5,65,66
C) 6,5,6,5
D) 6,5,6,6
Answer: A
Explanation:
According to scanf format parsing:
%c → ‘6’ (character ‘6’, ASCII code 54)
%c → ‘5’ (character ‘5’, ASCII code 53)
%d → 65 (integer 65, corresponding to ASCII character ‘A’)
%d → 66 (integer 66, corresponding to ASCII character ‘B’)
Variable assignments:
a = ‘6’ (character ‘6’)
b = ‘5’ (character ‘5’)
c = 65 (integer 65, but when output with %c it will convert to ‘A’)
d = 66 (integer 66, but when output with %c it will convert to ‘B’)
09
The output of the following program is
main()
{ char *p1,*p2;
char str1[50]=”xyz”,str2[]=”abcd”;;
p1=str2;
p2=”ABCD”;
strcpy(str+2,strcat(p1+2,p2+1));
printf(“%s”,str); }
A) xyabcAB
B) abcABz
C) ABabcz
D) xycdBCD
Answer: D
Explanation:
The main function of the program is to manipulate strings and pointers: defining character pointers p1, p2 and character array str initialized to “xyz”; p1 points to the string constant “abcd”, p2 points to “ABCD”; using strcat and strcpy for string concatenation and copying; finally printing the content of str.
strcat function: its function is to append the src string to the end of the dest string, overwriting the terminating null character ‘\0’ of dest, returning a pointer to dest, dest must be a modifiable character array, not a string constant.
10 (Common Mistake)
Read the following program:
# include <stdio.h>
main()
{ int a[]={1,2,3,4,5,6,7,8,9,10,11,12};
int*p=a+5,*q=NULL;
*q=*(p+5);
printf(“%d,%d \n”,*p,*q);
}
The output of the program is
A) Runtime error
B) 6,6
C) 6,11
D) 5,10
Answer: A
Explanation:
The reason for the error: q is initialized to NULL, dereferencing *q directly will cause a runtime error. The correction method: q should first point to a valid memory address.