This question already has answers here:
How do I properly compare strings in C?
(10 answers)
Closed 1 year ago.
I'm trying the following codes, but got unexpected result.
//the first element in argv[] is a
int main(int argc, char *argv[]) {
char a;
if (*argv == "a")
{
printf("a");
}
}
I got nothing after excution, so that means the condition *argv++ == "a" is false. So why?
In this if statement
if (*argv == "a")
there are compared two pointers. The first one is the pointer to the first character of the string pointed to by the pointer *argv and the second one is the pointer to the first character of the string literal "a".
As these two strings occupy different extents of memory then the comparison of the pointers always evaluates to logical false.
If you want to compare pointed strings then you need to use the standard C string function strcmp. For example
if ( strcmp( *argv, "a" ) == 0 )
If you want to compare first characters of the strings then you should write
if ( **argv == *"a" )
or just use the character 'a' instead of the string literal that is more natural
if ( **argv == 'a' )
Pay attention to that if *argv is not equal to NULL then this pointer points to the string that denotes the running program.
Maybe you actually mean the following if statement
if ( argc > 1 && strcmp( argv[1], "a" ) == 0 )
if you want to check whether the first command line argument is equal tp the string literal "a".
char *argv[] is an array of pointers to char.
// myapp.c - program to print out all command line parameters
#include <stdio.h>
int main (int argc, char *argv[]) {
int i ;
for (i = 0; i < argc; i++) {
printf ("arg %d = %s\n", i + 1, argv[i]) ;
}
return 0 ;
}
If one were to compile this code and run it at the command line like this:
myapp This That another
It would print out
arg 1 = myapp
arg 2 = This
arg 3 = That
arg 4 = another
So, let's say *argv points to 0x0800001000 (in RAM) and "a" is held at address 0x090000000 (in RAM). What you are doing is like saying:
if (0x0800001000 == 0x090000000) {
...
}
which is always going to be false (unless something really weird happens)
I'm not entirely sure what you want to do, but there are two possibilities
You want to compare the first string passed in on the command line to "a"
// do it like this
if (strcmp(argv[1],"a") == 0) {
...
}
You want to compare the first character of the first string passed in on the command line to 'a'
// do it like this
if (argv[1][0] == 'a') {
...
}
We use argv[1] because argv[0] is just the name of the executable
Related
I am writing a program in C for a basic calculator. I am trying to do this using what I have learned so far: printf() and scanf() functions. I am passing arguments into my program through the command line. I am assuming three arguments will be passed at a time which includes: first int, an operator, and the second int. I want to check if the second arg passed is an operator and then check if it's +,-,*... so on. Here is what I came up with:
int main(int argc, char **argv) {
scanf("%d %c %d", &a, &oper, &b);
if (oper != 43) {
printf("Error: Operator is not a +");
return(1);
}
}
So obviously, I have omitted a lot of the code and kept the relevant part. Here I am just checking if the oper is a +. The ASCII key is 43 so I thought this would work but no luck! Any ideas? (I would like to see if I can do this just with printf and scanf if possible)
EDIT: For example if 12 b 13 was entered, it should return the error above. Same goes for '10 +a 10' or '10 ++ 10'.
Firstly I would highly recommend looking at the man-pages for any C library function you come across, they have a lot of useful information. It seems like you are using scanf() improperly as it is not made to be used with command line arguments.
You can check for matches for a single character by comparing the argument like this:
if(argv[2][0] == '+') ...
(argv[0] is the program's file name).
If would would like to compare string you can use strcmp(). But for the operator example you can get away with just checking the first and second characters in the argument like this:
if(argv[2][0] == '+' && argv[2][0] == '\0') ...
What this does is compare the first two characters of the argument. It first checks for the '+' and then checks if that is the end of the string with by checking for the null terminator '\0'.
We can make the assumption that any argument has at least two characters, the visible character and a null terminator. Performing this on other strings has no guarantee of this however.
The other characters, specifically the numbers need to be converted from their respective ASCII values to integers. You can use atoi or strtol to do this, although atoi will most likely be easier for you.
As David C. Rankin pointed out, **argv is a double pointer which at a high level and in most cases you can treat as a double array. In C a string is actually just an array of type char, so what argv[2] is doing above is first accessing the third index of **argv, this is now de-referenced to a type char * where the string (char array) is located. This can then further be de-referenced by the [0] in argv[2][0] to look at the first char of the string.
Code example:
char **my_arrays = argv; // a array of arrays
char *array = *argv; // de-references to index 0 in argv
char *array = *(argv + 1); // de-references to index 1 in argv
char *array = argv[0]; // de-references to index 0 in argv
char *array = argv[1]; // de-references to index 1 in argv
char first_char = *(*argv) // the first char of the first array of argv
char first_char = *(argv[0]) // the same as above
char first_char = argv[0][0] // the same as above
A side note. All strings in C should end in a null terminator which can be represented by NULL, 0, or '\0' values. This will represent the end of the string and many C functions rely on this to know when to stop.
Also NULL is technically a C macro, but you don't need to treat it any differently than 0 because it literally just expands to 0.
It's char **argv. As Some programmer dude said, you should reread your book/tutorial.
scanf doesn't read arguments. It reads from stdin.
Arguments are of type char* and are stored in argv. To convert these arguments to integers, use atoi or strtol (preferably strtol). See this for more info.
If you want to read from stdin using scanf, that is fine, and what you have will work as long as you instead input the data into stdin, and not as command line arguments.
This question already has answers here:
How to properly compare command-line arguments?
(4 answers)
Closed 4 years ago.
So I have to write a program that contains two possible options depending on what the user chooses by entering either -i or -w. I'm very new to command line arguments in general and I have no idea how to do this. So far I have:
#include <stdio.h>
int main(int argc, char *argv[])
{
if(argc == -'i') {
puts("Test 1");
}
else if(argc == -'w') {
puts("Test 2");
}
return 0;
}
It doesn't print anything...
Any explanation is greatly appreciated. I'm just trying to understand the logic behind this.
First of all, you are comparing oranges with appels. argc stores the number of
arguments. Second, even if you used argv[1], the comparison would be still
wrong, because it would be comparing pointers, not the contents. Note that in C
a string is a sequence of characters that terminates to '\0'-terminating byte.
A char* variable only points to the start of that sequence. The == operator
checks for equality of values, in case of pointers (and string literals), it
compares whether both pointers point to the same location.
When you want to compare strings, you have to compare the strings themselves,
that means you have to compare the contents where the pointers are pointing to.
For strings you have to use the strcmp function like this:
#include <stdio.h>
int main(int argc, char *argv[])
{
if(argc != 2)
{
fprintf(stderr, "usage: %s option\n", argv[0]);
return 1;
}
if(strcmp(argv[1], "-i") == 0)
puts("Test 1");
else if(strcmp(argv[1], "-w") == 0)
puts("Test 2");
else {
fprintf(stderr, "Invalid option '%s'\n", argv[1]);
return 1;
}
return 0;
}
Note that it is important to check first that you have enough command line
arguments (the first if of my code). If the user didn't pass any argument,
argc will be 1 and argv[1] will point to NULL, which would yield undefined
behaviour if it is passed to strcmp. That's why I only do strcmp when I'm
100% sure that argv[1] is not NULL.
Also if you are coding for a POSIX system (linux, mac, etc), then I recommend using
getopt for parsing of the command line arguments.
You have to check argv[i] where i is the array number of the command line argument being put in, argv[0] would be the file name called upon and after that argv[1] would be the first statement, argv[2] the next and so on
argc means "argument count". meaning the number of arguments
argv is a 2-dimensional array. Strings are 1-dimensional character arrays in C. The second dimension comes from you having multiple String.
So if you want the first String argument, you would access it as follows:
argv[0]
You are also attempting to compare strings, which are more than 1 character long. You should use strcmp to compare strings in C. See How do I properly compare strings?
and if you want to compare equality, you would not use ==, == is for basic data types such as int or char.
argc represents the number of parameters that were passed in at the command line including the program name itself.
In c, a character, e.g., 'i' is an 8-bit number representing the ASCII code of the letter i. So your conditional statement if(argc == -'i') is actually checking whether -105 (105 is the ascii value of the letter i) is the number of arguments that was passed to your program.
To check whether the arguments passed in are "-i" or "-w" you need to perform string comparison, using the c library functions or your own algorithm, and you need to be comparing those strings to argv[i] (where i is the position of the parameter you're checking in the program invocation)
#include <stdio.h>
int main(int argc, char *argv[])
{
printf("The name of the program is %s\n", argv[0]);
if( strcmp(argv[1], "-i") == 0 ) { puts("Test 1"); }
if( strcmp(argv[1], "-w") == 0 ) { puts("Test 2"); }
return 0;
}
I have this code:
int main(int argc, char *argv[]) {
int num = *argv[1];
When I run the function in terminal with a parameter: for example, if I were to call ./main 17, I want num = 17. However, with this code, num = 49 (ASCII value for 1 because argv is an array of characters). How would I get it to read num = 17 as an int? Playing around with the code, I can get it to convert the parameter into an int, but it will still only read/convert the first value (1 instead of 17).
I'm new to C and the concept of pointers/pointers to arrays is still confusing to me. Shouldn't *argv[1] return the value of the second char in the array? Why does it read the first value of the second char in the array instead?
Thanks for help!
How do you convert parameters from char to int?
Can be done by a simple cast (promotion), but this isn't your case.
In your case *argv[] is array of pointer to char (You can use this for breaking down complex C declarations), meaning that argv[1] is the 2nd element in the array, i.e. the 2nd char* in the array, meaning *argv[1] is the first char in the 2nd char* in the array.
To show it more clearly, assume argv holds 2 string {"good", "day"}. argv[1] is "day" and *argv[1] is 'd' (note the difference in types - char vs char*!)
Now, you are left with the 1st char in your input string i.e. '1'. Its ascii is indeed 49 as, so in order to get it's "int" value you should use atoi like this:
int i = atoi("17");
BUT atoi gets const char * so providing it with 17 is a good idea while sending it a char would not. This means atoi should get argv[1] instead of *argv[1]
int main(int argc, char *argv[]) {
int num = atoi(argv[1]);
// not : int num = *argv[1]; --> simple promotion that would take the ascii value of '1' :(
// and not: int num = atoi(*argv[1]); --> argument is char
note: atoi is considered obsolete so you may want to use long int strtol(const char *str, char **endptr, int base) but for a simple example I preferred using atoi
Shouldn't *argv[1] return the value of the second char in the array?
Look at the signature:
int main(int argc, char *argv[])
Here, argv is an array ([]) of pointers (*) to char. So argv[1] is the second pointer in this array. It points to the first argument given at the command line. argv[0] is reserved for the name of the program itself. Although this can also be any string, the name of the program is put there by convention (shells do this).
If you just dereference a pointer, you get the value it points to, so *argv[1] will give you the first character of the first argument. You could write it as argv[1][0], they're equivalent. To get the second character of the first argument, you'd write argv[1][1].
An important thing to note here is that you can never pass an array to a function in C. The signature above shows an array type, but C automatically adjusts array types to pointer types in function declarations. This results in the following declaration:
int main(int argc, char **argv)
The indexing operator ([]) in C works in terms of pointer arithmetics: a[x] is equivalent to *(a+x). The identifier of an array is evaluated as a pointer to the first array element in most contexts (exceptions include the sizeof operator). Therefore indexing works the same, no matter whether a is an array or a pointer. That's why you can treat argv very similar to an array.
Addressing your "core" problem: You will always have strings in argv and you want numeric input, this means you have to convert a string to a number. There are already functions doing this. A very simple one is atoi(), you can use it like this:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
if (argc != 2)
{
// use program name in argv[0] for error message:
fprintf(stderr, "Usage: %s [number]\n", argv[0]);
return EXIT_FAILURE;
}
int i = atoi(argv[1]);
printf("Argument is %d.\n", i);
return EXIT_SUCCESS;
}
This will give you 0 if the argument couldn't be parsed as a number and some indeterminate value if it overflows your int. In cases where you have to make sure the argument is a valid integer, you could use strtol() instead (note it converts to long, not int, and it can handle different bases, so we have to pass 10 for decimal):
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
int main(int argc, char **argv)
{
if (argc != 2)
{
// use program name in argv[0] for error message:
fprintf(stderr, "Usage: %s [number]\n", argv[0]);
return EXIT_FAILURE;
}
errno = 0; // reset error number
char *endptr; // this will point to the first character not read by strtol
long i = strtol(argv[1], &endptr, 10);
if (errno == ERANGE)
{
fprintf(stderr, "This number is too small or too large.\n");
return EXIT_FAILURE;
}
else if (endptr == argv[1])
{
// no character was converted. This also catches the case of an empty argument
fprintf(stderr, "The argument was not a number.\n");
return EXIT_FAILURE;
}
else if (*endptr)
{
// endptr doesn't point to NUL, so there were characters not converted
fprintf(stderr, "Unexpected characters in number.\n");
return EXIT_FAILURE;
}
printf("You entered %ld.\n", i);
return EXIT_SUCCESS;
}
I'm new to C and the concept of pointers/pointers to arrays is still confusing to me.
In C strings are represented by null terminated ('\0') character arrays. Let's consider the following example:
char str[] = "Hello world!"
The characters would lie contiguous in memory and the usage of str would decay to a character pointer (char*) that points to the first element of the string. The address of (&) the first element taken by &str[0] would also point to that address:
| . | . | . | H | e | l | l | o | | W | o | r | l | d | ! | \0 | . | . | . |
^ ^
str null terminator
Shouldn't *argv[1] return the value of the second char in the array?
First of all in the argv is an array of character pointer char* argv[], so that it could be interpreted like an array of strings.
The first string argv[0] is the program name of the program itself and after that the arguments that are passed are coming:
argv[0] contains a pointer to the string: "program name"
argv[1] contains a pointer to the argument: "17"
If you dereference argv[1] with the use of * you get the first character at that address, here 1 which is 49 decimal in the Ascii code. Example:
p r ("program name")
^ ^
argv[0] (argv[0] + 1)
--------------------------------------------
1 7 ("17")
^ ^
argv[1] (argv[1] + 1)
How would I get it to read num = 17 as an int?
Check the number of passed arguments with argc which contains also the program name as one (read here more about argc and argv). If there are 2 you can use strtol() to convert argv[1] to the an integer. Use strtol() over atoi() because atoi() is considered to be deprecated because there is no error checking available. If atoi() fails it simply returns 0 as integer instead of strtol() that is setting the second argument and the global errno variable to a specific value.
The followig code will use the pointer that strtol() set the second argument to, to check for conversion errors. There are also overflow and underflow errors to check like it's described here on SO. Moreover you have to check if the returned long value would fit into an int variable if you want to store it into an int variable. But for simplicity I've left that out:
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char* argv[])
{
/* At least 1 argument passed? */
if (argc >= 2)
{
char* endptr;
long num = strtol(argv[1], &endptr, 10);
/* Were characters consumed? */
if (argv[1] != endptr)
{
printf("Entered number: %ld\n", num);
}
else
{
printf("Entered argument was not a number!\n");
}
}
else
{
printf("Usage: %s [number]!\n", argv[0]);
}
return 0;
}
Here's what you want to do:
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char *argv []) {
int num = atoi (argv[1]);
printf ("Here's what you gave me: %d", num);
return 0;
}
Here's the documentation for atoi ().
argv is an array of strings, so argv[x] points to a string. atoi () accepts an ASCII string as input and returns an int.
Bonus: This is still a bit unsafe. Try running this program without passing it a parameter and see what happens.
Also, you must take a look at the documentation for strtol (), which is a safe way of doing this.
I want to read argument and then compare it with a character:
int main (int argc, char *const argv[]) {
if (argv[1][1] == 'c') {
client();
}
else if (argv[1][1] == 's') {
server();
}
return 0;
}
It works if I type cc or dd, it also works if I type ccttttt. It's just taking the second character, but I'd like it to work only if I type c.
Just change argv[1][1] to argv[1][0] and remember in c/c++ all arrays start in '0'
argv[1] is a pointer to the first command-line argument.
argv[1][1] is the second character of that argument.
The first character is argv[1][0]. (But first check that argc >= 2, i.e., that there actually is a command-line argument.)
Of course this only checks a single character, so it doesn't distinguish between "c" and "cthulhu". If that's how you want to handle the argument, that's fine, but you might want to consider a different approach.
This question already has answers here:
Using the equality operator == to compare two strings for equality in C [duplicate]
(9 answers)
Closed 8 years ago.
In the following working code; instead of using *tofind, if I directly use the comparison
if(*argv[i] == "and")
it fails.
Why would that be?
/**
* Find index of the word "and"
* ./a.out alice and bob
*/
int main(int argc, char **argv) {
int i = 0;
char *tofind = "and";
while (argv[i] != NULL) {
if(*argv[i] == *tofind) {
printf("%d\n", i + 1);
break;
}
++i;
}
return 0;
}
if(*argv[i] == "and") shouldn't compile, I think you mean if (argv[i] == "and"), that would compare the pointers of the two, not the string content.
if (*argv[i] == *tofind) doesn't work as you expected either, it only compares the first character.
To compare strings, use strcmp():
if (strcmp(argv[i], tofind) == 0)
A "char*" is officially a pointer to a single char, for example to find points to a letter 'a'. You know that there are two more chars and a nul char, but officially it points to a single char.
Therefore, *argv[i] is the first character of an argument, and *tofind is always the letter 'a', so your code checks if the first character of an argument is an 'a'. Look at the strcmp function, which compares whole strings.
look at the type of
*argv[i] //its type is char
and of "and"
"and" //its type is const char * as it is decayed into pointer
so thats why you are unable to compare them.
while the type of
*tofind
is char and you can now compare the two.for more details see FAQs section 6.