question about open file in unix system

Hi,all
I use fopen to open a txt file in Unix , but it always fails (ipf =�@NULL ). The file 1.txt is already in the current directory.
I don' t know why. Can you give some advice?

Thanks,

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char *File1=NULL,*File2=NULL;
FILE *ifp=NULL,*ofp=NULL;

File1= "./1.txt";
File2= "./1-1.txt";

ifp=fopen(File1,"r");
ofp=fopen(File2,"w");

char File1[] = "./1.txt";

and the same for the other file.

Try two '/', I'm on windows at present and can't verifiy.

File1= ".//1.txt";
File2= ".//1-1.txt";

this works for me on UNIX

Code:

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#include <stdio.h>

int main()
{
    FILE *fp1, *fp2;
    char *filename1 = "./1.txt";
    char *filename2 = "./1-1.txt";
   
    if( (fp1 = fopen( filename1, "r" ) ) == NULL)
        printf("Unable to open file.\n");
    else printf("File %s is open for reading.\n", filename1);
   
    if( (fp2 = fopen( filename2, "w" ) ) == NULL)
        printf("Unable to open file.\n");
    else printf("File %s is open for writing.\n", filename2);
           

    return 0;
}

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hope this helps

Quote:

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printf("Unable to open file.\n");

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To print out error, try using the function perror(), is better.

Quote:

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Originally posted by Vber
To print out error, try using the function perror(), is better.

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Thanks.

I have used it, can you tell me how to know the decsription of the error?

errno =�@0x290010

It happens when fclose ends.

Post some of your code. Also, this is how it is supposed to look:

Code:

---

char *fn = "myfile.txt";

if ((fp = fopen(fn, "r")) == NULL)
{
  perror (fn);
  return (EXIT_FAILURE);
}

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