Command Line Arguments Questions and Answers

Q: What will be the output of the following C statement? (assuming the input is “cool brother in city”) printf("%s\n", argv[argc]);

Correct answer: (null)argv[argc] is guaranteed to be a null pointer, so the call is printf("%s\n", NULL).Passing a null pointer to %s is undefined behaviour under the standard, but the GNU C library detects it and prints the literal text (null) rather than dereferencing it - which is the behaviour this key reflects. Other implementations crash instead, so the code should not be written whatever library it happens to meet.

Q: What is the output of this C code? #include <stdio.h> void main() { int x = 0; int *ptr = &5; printf("%p\n", ptr); }

Correct answer: Compile time error& requires an lvalue - something with a storage location - and the literal 5 has none. It is a value, not an object, so the address-of operator has nothing to take the address of.The compiler rejects the expression outright. Assigning the literal to a variable first, and taking the address of that, is the only way to obtain a pointer to a 5; and a string literal is the special case that does have an address, being an array object in its own right.

Q: Which of the following is TRUE about argv?

Correct answer: It is an array of character pointersDeclared as char *argv[], argv is an array whose elements are char * - one pointer per argument, each addressing a null-terminated string.The second option describes char **argv read too literally: as a parameter the array does decay to a pointer to its first element, but the object itself is an array of pointers. "An array of strings" is loose talk, since C has no string type; what it has is an array of pointers to characters, which is precisely the first option.

Q: In Turbo C/C++ under DOS if we want that any wild card characters in the command-line arguments should be appropriately expanded, are we required to make any special provision?

Correct answer: YesUnder DOS the shell does not expand wildcards - a program invoked as prog *.c receives the literal string "*.c" and must expand it itself.Turbo C supplies that facility in a separate object module, WILDARGS.OBJ, which has to be linked in explicitly; only then does the runtime expand the pattern into matching filenames before main is entered. Unix-like shells do the expansion themselves before the program starts, so no such provision is needed there.

Q: What will be the output of the program (sample.c) given below if it is executed from the command line? cmd> sample 1 2 3 cmd> sample 2 2 3 cmd> sample 3 2 3 /* sample.c */ #include <stdio.h> int main(int argc, char *argv[]) { printf("%s\n", argv[0]); return 0; }

Correct answer: sampleargv[0] holds the name by which the program was invoked, and the command line here begins with sample, so that is what is printed - whatever arguments follow.The three invocations differ only in their arguments, and argv[0] is unaffected by them. Note that the exact contents are a platform convention: Unix-like systems generally give the string as typed, while Turbo C under DOS supplies the full uppercase path, which is what the companion question reflects.

Q: What will be the output of this program (prg_3.c)?#include <stdio.h> #include <stdlib.h> int main(int argc,char* argv[]) { int answer; answer=atoi(argv[1])+atoi(argv[2]); printf("%d",answer); return 0; }The program is run as:C:\TC\BIN>prg_3.exe 10 20

Correct answer: 30Command line arguments always arrive as strings, never as numbers. With the command line prg_3.exe 10 20, argv[1] is the two-character string "10" and argv[2] is "20". Adding those directly would add two pointers, which C does not even permit, so the program converts each with atoi - ASCII to integer, declared in <stdlib.h> - giving \[ 10 + 20 = 30. \] Two habits are worth taking from this. First, atoi reports no errors: given text that is not a number it simply returns 0, which is indistinguishable from a genuine "0", so production code uses strtol instead. Second, the program never checks argc before reading argv[1] and argv[2]; run with fewer arguments it would dereference a null pointer and crash. The command line had been lost from this record and is restored above.

Q: What is the output of this C code? #include <stdio.h> int main() { int i = 11; int *p = &i; foo(&p); printf("%d ", *p); } void foo(int *const *p) { int j = 10; *p = &j; printf("%d ", **p); }

Correct answer: Compile time errorRead the parameter from the identifier outwards: p is a pointer to a const pointer to int. So *p - the pointer being pointed at - is qualified const, and *p = &j attempts to modify it.The compiler rejects the assignment. The companion question uses int **const p, where the const attaches to p itself and *p remains modifiable - the same two words in the other order giving the opposite permission.

Q: What is the output of this C code? #include <stdio.h> int x = 0; void main() { int *const ptr = &x; printf("%p\n", ptr); ptr++; printf("%p\n ", ptr); }

Correct answer: Compile time errorint *const ptr declares a constant pointer to a modifiable int, so ptr itself may not be reassigned and ptr++ is rejected.Contrast const int *ptr, where the const protects the pointed-to value and the pointer may still be moved. The position of the qualifier relative to the * decides which is fixed, and reading from the identifier outwards makes it clear: ptr is a const pointer to int.

Q: If the different command line arguments are supplied at different times would the output of the following program change? #include <stdio.h> int main(int argc, char **argv) { printf("%d\n", argv[argc]); return 0; }

Correct answer: Noargv[argc] is guaranteed by the standard to be a null pointer, marking the end of the vector. That is true whatever arguments are supplied, so the value printed never changes.Note the further problem: the format is %d while the argument is a pointer, which is itself undefined behaviour and typically prints 0. The question turns on the guarantee at the end of the array - a program can therefore walk argv until it meets the null pointer without consulting argc at all.

21.

What is the second argument in command line arguments?

A.The number of command-line arguments the program was invoked with;
B.A pointer to an array of character strings that contain the arguments, one per string
C.Nothing
D.None of the mentioned

Answer: B

Correct answer: A pointer to an array of character strings that contain the arguments, one per string

The second parameter is argv. Declared char *argv[] it is an array of pointers, and as a parameter it decays to a pointer to its first element, which is how char **argv comes to be an equivalent declaration.

Each element addresses one null-terminated argument, with argv[0] naming the program and argv[argc] guaranteed null. The first parameter, argc, carries the count of those strings.

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22.

What will be the output of the following C statement? (assuming the input is “cool brother in city”)

printf("%s\n", argv[argc]);

A.(null)
B.City
C.In
D.Segmentation Fault

Answer: A

Correct answer: (null)

argv[argc] is guaranteed to be a null pointer, so the call is printf("%s\n", NULL).

Passing a null pointer to %s is undefined behaviour under the standard, but the GNU C library detects it and prints the literal text (null) rather than dereferencing it - which is the behaviour this key reflects. Other implementations crash instead, so the code should not be written whatever library it happens to meet.

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23.

What is the output of this C code?

#include <stdio.h>
    void main()
    {
        int x = 0;
        int *ptr = &5;
        printf("%p\n", ptr);
    }

A.5
B.Address of 5
C.Nothing
D.Compile time error

Answer: D

Correct answer: Compile time error

& requires an lvalue - something with a storage location - and the literal 5 has none. It is a value, not an object, so the address-of operator has nothing to take the address of.

The compiler rejects the expression outright. Assigning the literal to a variable first, and taking the address of that, is the only way to obtain a pointer to a 5; and a string literal is the special case that does have an address, being an array object in its own right.

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24.

Which of the following is TRUE about argv?

A.It is an array of character pointers
B.It is a pointer to an array of character pointers
C.It is an array of strings
D.None of above

Answer: A

Correct answer: It is an array of character pointers

Declared as char *argv[], argv is an array whose elements are char * - one pointer per argument, each addressing a null-terminated string.

The second option describes char **argv read too literally: as a parameter the array does decay to a pointer to its first element, but the object itself is an array of pointers. "An array of strings" is loose talk, since C has no string type; what it has is an array of pointers to characters, which is precisely the first option.

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25.

In Turbo C/C++ under DOS if we want that any wild card characters in the command-line arguments should be appropriately expanded, are we required to make any special provision?

A.Yes
B.No

Answer: A

Correct answer: Yes

Under DOS the shell does not expand wildcards - a program invoked as prog *.c receives the literal string "*.c" and must expand it itself.

Turbo C supplies that facility in a separate object module, WILDARGS.OBJ, which has to be linked in explicitly; only then does the runtime expand the pattern into matching filenames before main is entered. Unix-like shells do the expansion themselves before the program starts, so no such provision is needed there.

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26.

What will be the output of the program (sample.c) given below if it is executed from the command line?
cmd> sample 1 2 3
cmd> sample 2 2 3
cmd> sample 3 2 3

/* sample.c */
#include <stdio.h>

int main(int argc, char *argv[])
{
    printf("%s\n", argv[0]);
    return 0;
}

A.sample 3 2 3
B.sample 1 2 3
C.sample
D.Error

Answer: C

Correct answer: sample

argv[0] holds the name by which the program was invoked, and the command line here begins with sample, so that is what is printed - whatever arguments follow.

The three invocations differ only in their arguments, and argv[0] is unaffected by them. Note that the exact contents are a platform convention: Unix-like systems generally give the string as typed, while Turbo C under DOS supplies the full uppercase path, which is what the companion question reflects.

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27.

What will be the output of this program (prg_3.c)?

#include <stdio.h>
#include <stdlib.h>
int main(int argc,char* argv[])
{
    int answer;

    answer=atoi(argv[1])+atoi(argv[2]);
    printf("%d",answer);
    return 0;
}

The program is run as:

C:\TC\BIN>prg_3.exe 10 20

A.50
B.60
C.30

Answer: C

Correct answer: 30

Command line arguments always arrive as strings, never as numbers. With the command line prg_3.exe 10 20, argv[1] is the two-character string "10" and argv[2] is "20". Adding those directly would add two pointers, which C does not even permit, so the program converts each with atoi - ASCII to integer, declared in <stdlib.h> - giving \[ 10 + 20 = 30. \] Two habits are worth taking from this. First, atoi reports no errors: given text that is not a number it simply returns 0, which is indistinguishable from a genuine "0", so production code uses strtol instead. Second, the program never checks argc before reading argv[1] and argv[2]; run with fewer arguments it would dereference a null pointer and crash. The command line had been lost from this record and is restored above.

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28.

What is the output of this C code?

#include <stdio.h>
    int main()
    {
        int i = 11;
        int *p = &i;
        foo(&p);
        printf("%d ", *p);
    }
    void foo(int *const *p)
    {
        int j = 10;
        *p = &j;
        printf("%d ", **p);
    }

A.Compile time error
B.10 10
C.Undefined behaviour
D.10 11

Answer: A

Correct answer: Compile time error

Read the parameter from the identifier outwards: p is a pointer to a const pointer to int. So *p - the pointer being pointed at - is qualified const, and *p = &j attempts to modify it.

The compiler rejects the assignment. The companion question uses int **const p, where the const attaches to p itself and *p remains modifiable - the same two words in the other order giving the opposite permission.

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29.

What is the output of this C code?

#include <stdio.h>
    int x = 0;
    void main()
    {
        int *const ptr = &x;
        printf("%p\n", ptr);
        ptr++;
        printf("%p\n ", ptr);
    }

A.0 1
B.Compile time error
C.0xbfd605e8 0xbfd605ec
D.0xbfd605e8 0xbfd605e8

Answer: B

Correct answer: Compile time error

int *const ptr declares a constant pointer to a modifiable int, so ptr itself may not be reassigned and ptr++ is rejected.

Contrast const int *ptr, where the const protects the pointed-to value and the pointer may still be moved. The position of the qualifier relative to the * decides which is fixed, and reading from the identifier outwards makes it clear: ptr is a const pointer to int.

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30.

If the different command line arguments are supplied at different times would the output of the following program change?

#include <stdio.h>

int main(int argc, char **argv)
{
    printf("%d\n", argv[argc]);
    return 0;
}

A.Yes
B.No

Answer: B

Correct answer: No

argv[argc] is guaranteed by the standard to be a null pointer, marking the end of the vector. That is true whatever arguments are supplied, so the value printed never changes.

Note the further problem: the format is %d while the argument is a pointer, which is itself undefined behaviour and typically prints 0. The question turns on the guarantee at the end of the array - a program can therefore walk argv until it meets the null pointer without consulting argc at all.

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