Problems using main(int argc,char *argv[])

Question

#include<stdio.h>
#include<string.h>
int main(int argc,char *argv[])
{
    char string[]="#";
    printf("%s\n",argv[1]);
    printf("%s\n",argv[2]);
    printf("%s\n",argv[3]);
    strcat(argv[1],string);
    printf("argv[1] is %s\n",argv[1]);
    printf("argv[2] is %s\n",argv[2]);
    printf("argv[3] is %s\n",argv[3]);
    return 0;
}

when I use strcat() to add something to the end of the argv[1],argv[2] will be lost( strlen(argv[2]) changes to 0 after use strcat ).But argv[3] has no change .Why???

possibly duplicate of stackoverflow.com/questions/963493/… — Arun
– Arun, Commented Apr 10, 2013 at 2:35
@Thomas This code is just to solve my question with argv[].It can do nothing meaningful. — dsfa24
– dsfa24, Commented Apr 10, 2013 at 3:18

Havenard · Accepted Answer · 2013-04-10 02:49:30Z

2

You cannot append stuff directly to argv[] because each argv[] is allocated serialized and to a space in memory that can only hold its original size.

To understand what is happening, imagine argv is something like this:

char buffer[] = "./program\0one\0two\0three\0";
char *argv[4] = { &buffer[0], &buffer[10], &buffer[14], &buffer[18] };

So as you can see, if you write something after "one" you will overwrite "two" and break the whole thing because they are serialized in memory.

To workaround this, you have to copy each argv[] to a bigger buffer where you can safely make the modifications. For instance:

char buf[1024];
strcpy(buf, argv[1]);
strcat(buf, string);
printf("argv[1] is %s\n", buf);

answered Apr 10, 2013 at 2:49

Havenard

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Comments

mvarshney · Accepted Answer · 2013-04-10 02:37:28Z

2

Remember that "strings" in C are zero terminated arrays of characters. Suppose the original four arguments were "one", "two", "three" and "four"; then these four arguments are stored in memory as

one\0two\0three\0four\0

when you appended # after the argv[1] (strcat will add a \0 also), the memory content becomes:

one\0two#\0hree\0four\0

you can now verify that argv[2] points to the \0 char, hence it is a empty string. argv[3] still points correctly, and it remains intact.

answered Apr 10, 2013 at 2:37

mvarshney

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Comments

Martin Beckett · Accepted Answer · 2013-04-10 02:29:22Z

0

You aren't supposed to modify argv[] = the effects are undefined

answered Apr 10, 2013 at 2:29

Martin Beckett

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4 Comments

Arun Over a year ago

I think so too, but is there any reference?

paxdiablo Over a year ago

Actually, you can modify argv[whatever] (the pointer). Modifying what the pointer points to is not so kosher.

Arun Over a year ago

@paxdiablo: So, is it like char const * argv[]?

Martin Beckett Over a year ago

No it's not necessarily const - the standard doesn't say what it has to be. Just that the memory isn't yours so you shouldn't change it.

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