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Jun 7th, 2005

makes pointer from integer without a cast

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Hellow I'm just started whit C.

I'm usig gcc and get this error:

C Syntax (Toggle Plain Text)
test.c: In function 'main':
test.c:8: warning: passing argument 1 of 'strlen' makes pointer from integer without a cast
.....
test.c: In function 'main':
test.c:8: warning: passing argument 1 of 'strlen' makes pointer from integer without a cast
.....

This is my code:

C Syntax (Toggle Plain Text)
#include <stdio.h>
#include <string.h>
 
int main(int argc, char *argv[]){
	int arg = atoi(argv[argc-1]);
	int relay;
	int i;
	for(i = 0; i < strlen(arg); i++)
	{
			if (strcmp(arg, "1") == 0){ 
				relay = 1;
		}else{
			if (strcmp(arg, "2") == 0){ 
				relay = relay + 2;
		}else{
			if (strcmp(arg, "3") == 0){ 
				relay = relay + 4;
		}else{
			if (strcmp(arg, "4") == 0){ 
				relay = relay + 8;
		}else{
			if (strcmp(arg, "5") == 0){ 
				relay = relay + 16;
		}else{
			if (strcmp(arg, "6") == 0){ 
				relay = relay + 32;
		}else{
			if (strcmp(arg, "7") == 0){ 
				relay = relay + 64;
		}else{
			if (strcmp(arg, "8") == 0){ 
				relay = relay + 128;
			}
			printf("relay %d/n", relay);	
		}}}}}}}
	}
}
#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[]){
	int arg = atoi(argv[argc-1]);
	int relay;
	int i;
	for(i = 0; i < strlen(arg); i++)
	{
			if (strcmp(arg, "1") == 0){ 
				relay = 1;
		}else{
			if (strcmp(arg, "2") == 0){ 
				relay = relay + 2;
		}else{
			if (strcmp(arg, "3") == 0){ 
				relay = relay + 4;
		}else{
			if (strcmp(arg, "4") == 0){ 
				relay = relay + 8;
		}else{
			if (strcmp(arg, "5") == 0){ 
				relay = relay + 16;
		}else{
			if (strcmp(arg, "6") == 0){ 
				relay = relay + 32;
		}else{
			if (strcmp(arg, "7") == 0){ 
				relay = relay + 64;
		}else{
			if (strcmp(arg, "8") == 0){ 
				relay = relay + 128;
			}
			printf("relay %d/n", relay);	
		}}}}}}}
	}
}

What is I doing wrong?

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stage

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Jun 7th, 2005

Re: makes pointer from integer without a cast

A couple of things:

First, you need to include "stdlib.h" for atoi. (Note, I think atoi is not standard, consider sprintf)

Second, you are trying to get the length of an integer ( strlen(arg). I'm not sure what you're trying to do. Then, you are trying to string compare integers (strcmp( arg, "8"), etc.

Please explain what the program is supposed to do and we'll tell you how to change it.

winbatch

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Jun 7th, 2005

Re: makes pointer from integer without a cast

The program is supposed to take the user argument and transform it to anothe number. Like if the user write 3 the number will be 4 or if the user write 12 the number will be 3 or if the user write 12345678 the number will be 255, got it?

stage

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Jun 7th, 2005

Re: makes pointer from integer without a cast

OK, but why the for loop? You only want the user input once correct?

winbatch

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Jun 7th, 2005

Re: makes pointer from integer without a cast

Yes, I just wante the user input once, but I don't see how I could do it whitout the loop.

stage

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Jun 7th, 2005

Re: makes pointer from integer without a cast

Anyone know how I can do this:
I want to take the user argument (argv[argc-1]), which is a number in one or several digts (1 to 8 digts) and transform it to another number.... Accordig to this table:
1 = 1
2 = 2
3 = 4
4 = 8
5 = 16
6 = 32
7 = 64
8 = 128

So if the user write 46 the number will be 40 (8+32) or if the user write 137 the number will be 69 (1+4+64)

stage

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Jun 7th, 2005

Re: makes pointer from integer without a cast

A different approach...

C Syntax (Toggle Plain Text)
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
 
int main(int argc, char *argv[])
{
   int relay = 0;
   int i;
   for ( i = 0; argv[1][i]; ++i )
   {
      int x = 1 << (argv[1][i] - '1 ');
      printf("x = %d\n", x);
      relay += x;
   }
   printf("relay = %d\n", relay);
   return 0;
}
 
/* my output
C:\Test>test 46
x = 8
x = 32
relay = 40
 
C:\Test>test 137
x = 1
x = 4
x = 64
relay = 69
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
   int relay = 0;
   int i;
   for ( i = 0; argv[1][i]; ++i )
   {
      int x = 1 << (argv[1][i] - '1 ');
      printf("x = %d\n", x);
      relay += x;
   }
   printf("relay = %d\n", relay);
   return 0;
}

/* my output
C:\Test>test 46
x = 8
x = 32
relay = 40

C:\Test>test 137
x = 1
x = 4
x = 64
relay = 69
*/

Dave Sinkula

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Jun 7th, 2005

Re: makes pointer from integer without a cast

Thanks, but I can't get your code to work, first i get this warning in gcc:

C Syntax (Toggle Plain Text)
test.c:11:34: warning: multi-character character constant
test.c:11:34: warning: multi-character character constant

Then I try your code and get:

C Syntax (Toggle Plain Text)
[stage@hal test]$ ./test 46
x = 1048576
x = 4194304
relay = 5242880
[stage@hal test]$ ./test 137
x = 131072
x = 524288
x = 8388608
relay = 9043968
[stage@hal test]$ ./test 46
x = 1048576
x = 4194304
relay = 5242880
[stage@hal test]$ ./test 137
x = 131072
x = 524288
x = 8388608
relay = 9043968

stage

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Jun 7th, 2005

Re: makes pointer from integer without a cast

Quote originally posted by stage ...

Thanks, but I can't get your code to work, first i get this warning in gcc:
C Syntax (Toggle Plain Text)
test.c:11:34: warning: multi-character character constant
test.c:11:34: warning: multi-character character constant

(Toggle Plain Text)

int x = 1 << (argv[1][i] - '1 ');

Hmm. How'd that extra space get in there? It should be like this.

C Syntax (Toggle Plain Text)
int x = 1 << (argv[1][i] - '1');
int x = 1 << (argv[1][i] - '1');

I don't know if that solves the issue. But do you see generally what the code was trying to do?

That is, take each character of the number entered. Subtracting '1' will from '1' - '8' will give a value from 0 - 7. Left shifting a 1 by this amount should calculate a value like your table. The loop just sums 'em up.

Dave Sinkula

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long time no c

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Jun 7th, 2005

Re: makes pointer from integer without a cast

Yes, it solved it. Thanks!

stage

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