getting segmentation fault if I dont pass any command line parameters

[doni]

Hi,

I am a beginner to C and I wrote a program that takes arguments and
prints it in hex. I am getting a segmentation fault if I dont pass any
arguments instead I want it to display the Usage message that I have.

Here is my simple program.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char * argv[])
{
char buffer[100];
char ch, * string;
int i, a=0;

if (argc < 1) {
printf("Usage: test <IPv4 Address> \n");
}
else {
string = argv[1];
strcpy(buffer, string);
a = strlen(buffer);
printf("a value is: %d", a);
for (i = 0; i < a; i++)
{
printf("Value is: %02x\n", buffer);
//printf("Value is: %X\n", buffer);
}
}
}

Thanks,
doni

 

[Ben Pfaff]

if (argc < 1) {
printf("Usage: test <IPv4 Address> \n");
}

argc is almost always at least 1 in a hosted environment, because
argv[0] is used to contain the program's name. You should be
testing whether argc is less than 2.

 

[CBFalconer]

Hi,

I am a beginner to C and I wrote a program that takes arguments and
prints it in hex. I am getting a segmentation fault if I dont pass any
arguments instead I want it to display the Usage message that I have.

Here is my simple program.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char * argv[])
{
char buffer[100];
char ch, * string;
int i, a=0;

if (argc < 1) {
printf("Usage: test <IPv4 Address> \n");
}
else {
string = argv[1];
strcpy(buffer, string);
a = strlen(buffer);
printf("a value is: %d", a);
for (i = 0; i < a; i++)
{
printf("Value is: %02x\n", buffer);
//printf("Value is: %X\n", buffer);
}
}
}

Try: "if (argc < 2) ...".

When printing an address (i.e. a) use the %p specifier, and cast
the address to void*.

 

[Army1987]

doni wrote: [...]

int i, a=0; [...]
printf("a value is: %d", a);

[...]
When printing an address (i.e. a) use the %p specifier, and cast
the address to void*.

Huh? What address?

 

[doni]

if (argc < 1) {
printf("Usage: test <IPv4 Address> \n");
}

argc is almost always at least 1 in a hosted environment, because
argv[0] is used to contain the program's name. You should be
testing whether argc is less than 2.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p\
);}return 0;}

Thanks, Ben.

doni

 

[CBFalconer]

doni wrote:
[...]

int i, a=0; [...]
printf("a value is: %d", a);

[...]
When printing an address (i.e. a) use the %p specifier, and cast
the address to void*.

Huh? What address?

Wasn't any. I read too quickly and carelessly.