main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}
Answers were Sorted based on User's Feedback
Answer / susie
Answer :
1 2
Explanation:
The sizeof() operator gives the number of bytes
taken by its operand. P is a character pointer, which needs
one byte for storing its value (a character). Hence
sizeof(*p) gives a value of 1. Since it needs two bytes to
store the address of the character pointer sizeof(p) gives 2.
Is This Answer Correct ? | 16 Yes | 7 No |
Answer / jha334201553
sizeof(*p) = sizeof(char) = 1
sizeof(p) = sizeof(void *)
I don't know the value of sizeof(p) .In deferent system the
value is deferent.In DOD, it's 2. int 32bits winNT, it's 4.
in 64bits WinNT, It's 8
Is This Answer Correct ? | 6 Yes | 0 No |
Answer / dilberphant
The results are indeterminate. The program code is in error.
1) The variadic function printf() requires an in-scope
prototype. No prototype for printf() has been provided.
2) main() is (by definition) a function that returns an
integer value. It is unclear which version of the C language
this program is intended to conform to, and for most
versions of the language, main() is required to include a
return <value>;
for some integer <value>
3) In a hosted environment, main() accepts either two
arguments (an int, and a char *[]) or none. Thus, either
main(int argc, char *argv[])
or
main(void)
are acceptable
4) The size of a pointer is dependant on operating platform
and C compiler implementation. The C language does not
define a "correct" value for sizeof (char *), and thus /any/
value for sizeof (char *) is acceptable (with the above
caveats about platform and compiler). The value is
unpredictable at a theoretical level.
Is This Answer Correct ? | 2 Yes | 1 No |
void func1(int (*a)[10]) { printf("Ok it works"); } void func2(int a[][10]) { printf("Will this work?"); } main() { int a[10][10]; func1(a); func2(a); } a. Ok it works b. Will this work? c. Ok it worksWill this work? d. None of the above
#include<stdio.h> main() { struct xx { int x=3; char name[]="hello"; }; struct xx *s=malloc(sizeof(struct xx)); printf("%d",s->x); printf("%s",s->name); }
void main() { char far *farther,*farthest; printf("%d..%d",sizeof(farther),sizeof(farthest)); }
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main() { char str1[] = {‘s’,’o’,’m’,’e’}; char str2[] = {‘s’,’o’,’m’,’e’,’\0’}; while (strcmp(str1,str2)) printf(“Strings are not equal\n”); }
How can i find first 5 natural Numbers without using any loop in c language????????
main() { printf("%x",-1<<4); }
main() { int i =10, j = 20; clrscr(); printf("%d, %d, ", j-- , --i); printf("%d, %d ", j++ , ++i); } a. 20, 10, 20, 10 b. 20, 9, 20, 10 c. 20, 9, 19, 10 d. 19, 9, 20, 10
write a program to count the number the same (letter/character foreg: 's') in a given sentence.
Write a program using one dimensional array to assign values and then display it on the screen. Use the formula a[i]=i*10 to assign value to an element.
1 Answers Samar State University,
{ int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }
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