I am running the following program and got errors:
First-chance exception at 0x0f32d440 (msvcr100d.dll) in c.exe: 0xC0000005: Access violation reading location 0x00000000.
Unhandled exception at 0x772815de in c.exe: 0xC0000005: Access violation reading location 0x00000000.
The program '[9048] c.exe: Native' has exited with code -1073741510 (0xc000013a).
Here is the code
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[], char *env[]) //char *argv[]
{
int i;
printf("These are the %d command- line arguments passed to main:\n\n", argc);
if(strcmp(argv[1],"123")==0)
{
printf("success\n");
}
else
for(i=0; i<=argc; i++)
//if(strcmp(argv[1],"abc")==0)
printf("argv[%d]:%s\n", i, argv[i]);
/*printf("\nThe environment string(s)on this system are:\n\n");
for(i=0; env[i]!=NULL; i++)
printf(" env[%d]:%s\n", i, env[i]);*/
system("pause");
}
The problem should be with the strcmp function but I dont know how to solve it.
Could anyone help?
You have (at least) two problems.
The first is doing this:
if(strcmp(argv[1],"123")==0)
without first checking that argc >= 2.
The second is this:
for(i=0; i<=argc; i++)
in that you should be processing arguments 0 thru argc - 1 inclusive. What that loop does is process arguments 0 through argc and argv[argc] is always NULL.
The following program illustrates one way to fix this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main (int argc, char *argv[]) {
int i;
printf ("These are the %d command-line argument(s) passed to main:\n", argc);
if ((argc >= 2) && (strcmp (argv[1], "123") == 0)) {
printf (" success\n");
} else {
for (i = 0; i < argc; i++) {
printf (" argv[%d] = [%s]\n", i, argv[i]);
}
}
return 0;
}
You can see that the comparison with "123" is only done after ensuring that argv[1] is populated correctly. In addition, the loop has been changed to exclude argv[argc] since that's not one of the arguments. A transcript follows:
pax> testprog
These are the 1 command-line argument(s) passed to main:
argv[0] = [testprog]
pax> testprog 123
These are the 2 command-line argument(s) passed to main:
success
pax> testprog a b c
These are the 4 command-line argument(s) passed to main:
argv[0] = [testprog]
argv[1] = [a]
argv[2] = [b]
argv[3] = [c]
for(i=0; i<=argc; i++) should be for(i=0; i<argc; i++).
C/C++ arrays are 0 to n-1. You are running 1 spot off the end of the array.
Related
When I was making my program I got an error type of thing called Segmentation Fault.
#include <cs50.h>
#include <string.h>
#include <stdio.h>
int main(int argc, string argv[])
{
char i = strlen(argv[2]);
if (argc == 2)
{
printf("%i %s %hhd", argc, argv[2], i);
}
}
I run this program using these commands
make substitution
and then
./substitution abcdefghijklmnopqrstuvwxyz
In this we have to add a 26 word key which in the above line is a to z.
Please help if you know to solve
If you invoke your program with:
./substitution abcdefghijklmnopqrstuvwxyz
argc will be 2, and argv will be an array of 3 pointers:
argv[0] points to the string ./substitution, argv[1] points to the string abcdefghijklmnopqrstuvwxyz, and argv[2] is NULL. If you attempt to compute the length of NULL by calling strlen(argv[2]), that is an error. You must not pass NULL to strlen. I think your error is simply mis-indexing the argv array. Arrays in C are zero based. If you want to compute the length of the first argument, you want to work with argv[1], not argv[2]:
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
int
main(int argc, char *argv[])
{
int rc = EXIT_FAILURE;
if( argc > 1 ){
size_t i = strlen(argv[1]);
printf("%i %s %zd\n", argc, argv[1], i);
rc = EXIT_SUCCESS;
}
return rc;
}
You got hit with segfault because the number of required arguments specified was done poorly. The main problem is here:
if (argc == 2)
The number of actual arguments passed by the user is equal to the number of required arguments minus 11. It should be:
int main(int argc, char *argv[]) {
if (argc != 3) {
// Number of arguments are either greater or lesser than 3
// Handle the error
}
// Ok...
}
You can safely use the arguments now.
1. The calling command for the program, for example ./a.out is also counted as an argument (argv[0]).
This question already has answers here:
How to print argv arguments from main function in C?
(7 answers)
Closed 2 years ago.
I have this c program right here:
#include <stdio.h>
#include <stdlib.h>
#include <Windows.h>
int main(int argc, char* argv[]) {
int i;
int forparam = sizeof(argv);
if (argc > 1) {
for (i = 1; i < forparam; i++) {
if (i == 1) {
printf("%s", argv[i]);
}
if (i > 1) {
printf(" %s", argv[i]);
}
}
}
else {
printf("One argument expected\n");
}
}
You type the name of the file in the Win10 cmd and a parameter and it prints out the parameter onto the cmd. At least that's what it's supposed to do. It can't take up more than 3 parameters and if you enter 1 or 2 parameters then it prints them out but with a "(null)" in the end. How can I make it to work? Because a space indicates another parameter it ignores it that's why I need all this code. Your help would be immensely appreciated!
sizeof(argv) gives you the size in bytes of the argv pointer object, not the number of elements it points to.
argc gives you the number of parameters on the command line, where argv[0] is the command used to invoke the program:
int main( int argc, char **argv )
{
printf( "Command used to invoke program: %s\n", argv[0] );
for ( int i = 1; i < argc; i++ )
printf( "parameter %d: %s\n", i, argv[i] );
}
You don't need forparam, you have everything you need with argc:
int main(int argc, char* argv[]) {
if (argc > 1) {
int i;
for (i = 1; i < argc; i++) {
if (i == 1) {
printf("%s", argv[i]);
} else if (i > 1) {
printf(" %s", argv[i]);
}
}
return EXIT_SUCCESS;
} else {
printf("One argument expected\n");
return EXIT_FAILURE;
}
}
Moreover, sizeof argv returns the size of the pointer argv, not the number of elements in argv (which is not known to the compiler).
Notice that sizeof(argv) is computed at compile-time, and is maybe 4 (bytes) or probably 8 (bytes). I suppose it is 8.
Then your for loop is executed with i=1, then with i=2 up to i=7
You probably want for (i = 1; i < argc; i++)
Read Modern C and see this C reference for more.
Take inspiration from the C source code of existing free software, in particular GNU coreutils.
This segment of code is meant to check if a user has entered only one numeric command-line argument, and return an error code of "1" if this is not the case. I have the code set up so that it first checks if argc is anything other than 2. Unfortunately, I am still receiving Segmentation Faults if no command line argument is entered, and I'm not sure why this code doesn't catch a null amount of command line arguments.
I tried moving the "if (argc !=2)" formula above the entire "for" statement to try and catch the command line argument issue right from the beginning, but I received the same result.
My question is, why am I receiving a Segmentation Fault when no command line argument is provided, and what am I missing to ensure the program doesn't Seg Fault with no command line argument?
Due to course policy, I will only be providing the segment of code in question.
#include <unistd.h>
#include <cs50.h>
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, string argv[])
{
// add'l variables //
int k = atoi(argv[1]);
for (int i = 0; i < strlen(argv[1]); i++)
{
if (argc != 2)
{
printf("Please enter only 1 command-line argument.\n");
return 1;
}
else if (!isdigit(argv[1][i]))
{
printf("Usage: ./caesar key\n");
return 1;
}
}
// add'l code //
Error as shown in Terminal
You must first check the argc before using the argv[1], because argv[1] may not have a valid pointer if argc < 2. A corrected version of your code could be like that:
#include <stdio.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
int i = 0;
if (argc != 2) {
printf("Please enter only 1 command-line argument.\n");
return 1;
}
while (isdigit(argv[1][i]))
++i;
if (argv[1][i] != '\0') {
printf("Usage: ./caesar key\n");
return 1;
}
return 0;
}
I am trying to build an echo-like command which I named code at xv6. The problem is that if:
Input:
$code Hello World
Output:
user space:Hello user space:World
While the correct output should be:
user space: Hello World
Can somebody help me?
int
main(int argc, char *argv[])
{
int i;
if(argc <= 1){
printf(1," %s user space:", argv[0]);
exit();
}
for(i = 1; i < argc; i++){
printf(1, " print in user space:%s",argv[i]);
}
printf(1, "\n");
exit();
}
program name is also passed as argument to your main function so you have three(as hello and world are considered separate arguments) arguments here .there is several issues in your code ,as you have two arguments first if statement will be false ,and you are not printing correctly in for loop ,following might be helpful :
int main(int argc, char *argv[])
{
int i;
printf("number of arguments : %d ",argc);
//argv array index count from 0
for(i = 0; i < argc; i++){
printf("argument number %d : %s",i,argv[i]); // issue in your code
}
printf("\n");//issue in your code
exit(0);
}
It's possible to use the write command so that you can output to stdout. Between each command line argument an additional space needs to be written.
#include <stdio.h>
#include <string.h>
#include <unistd.h>
int main(int argc, char *argv[]) {
int i;
write(1, "user space: ", 12);
for (i = 1; i < argc; i++) {
size_t len = strlen(argv[i]);
write(1, argv[i], len);
write(1, " ", 1);
}
write(1, "\n", 1);
return 0;
}
The problematic part in your code is here:
for(i = 1; i < argc; i++){
printf(1, " print in user space:%s",argv[i]); //" print in user space:" is printed
//for each iteration
}
As others pointed out, you should rather do:
printf(" print in user space:\n");
for(i = 1; i < argc; i++){
printf("%s",argv[i]);
}
That said, a few more remarks for you:
I never seen a printf that takes an integer as first argument, I assume that the first parameter is not important/is a typo
I am not an expert unix developer, but you can just use "return 0" (or "return 1") to quit your program. You will also be able to return an error code this way, as most unix programs do
If you decide to use printf, be aware that it comes with a nasty security exploit which can expose the caller's stack to a malicious user. Make sure your input is well formed or use a different function if possible
How does one create a function that takes another command as an argument and executes that command. For example say I wanted to do
./func1 cat /etc/motd
and have it execute the cat command on the file /etc/motd so that it would print the message of the day. If someone could show me that would be much appreciated!
EDIT: I cannot use the system() call as later on I have to program a basic command shell. I just need to know how to execute commands so that when a user types in cat foo.txt it executes the command and displays the file. I guess what I'm trying to say is, how do you use execve()? What arguments go inside it?
Use you can use the system function.
Example:
system("cat foo.txt");
Will run this:
cat foo.txt
You could do something like that:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
size_t command_length = 0;
if(argc < 2)
return 0;
for(int i = 1; i < argc; i++)
command_length += strlen(argv[i]);
command_length += argc - 2; // spaces between words
command_length++; // terminator '\0'
char command[command_length];
memset(command, 0, sizeof(command));
for(int i = 1; i < argc; i++) {
strcat(command, argv[i]);
if(i < argc - 1)
command[strlen(command)] = ' ';
}
system(command);
}
It first determines the length of all command line parameters. After that it concatenates all command line parameters and inserts a space between each. Last but not least it calls the system() function with this string.
You need to use a C11 compiler with VLA support.
Here is a version without system():
#include <string.h>
#include <unistd.h>
#define MAX 1024
int main(int argc, char *argv[]) {
char buf[MAX] = "/usr/bin/";
size_t len = MAX - strlen(buf) + 1;
if(argc < 2)
return 0;
strncat(buf, argv[1], len);
execve(buf, argv + 1, NULL);
return 0;
}
This program only works under Linux. Unfortunately the execve() expects an absolute path. I assumed that the executable is located under /usr/bin. Additional work is necessary if that's not the case. For example you would have to examine the $PATH environment variable.