CODEWITHSUNDEEP
c
PGOnTheGo
PGOnTheGo

Reputation: 805

Segmentation fault while trying to print contents of char*

I have the following piece of code:

    FILE *fpa;
    fpa = fopen(argv[2], "r");
    if (fpa == NULL) {
     printf("Error: could not open seqA file!\n");
     exit(0);
    }

    unsigned int N_a;
    fscanf(fpa, "%d\n", &N_a);
    char *seq_a = malloc((N_a+1) * sizeof(char *));
    strcpy(seq_a,"");
    fscanf(fpa, "%s\n", seq_a);
    fclose(fpa);

    for(i=0;i<N_a;i++)
      printf("%s", seq_a[i]); ---> SEG FAULT
    printf("\n");

I am getting a segmentation fault at the printf statement. argv[2] is a file whose contents is:
5
ABCBB

Any idea where i might be making a mistake.

Upvotes: 0

Views: 1465

Answers (2)

Jim Buck
Jim Buck

Reputation: 20726

Since you are printing characters, your printf should use format specifier %c.. not %s. %s expects a pointer, so it's treating a character value as a location in memory.

Upvotes: 0

P.P
P.P

Reputation: 121387

char *seq_a = malloc((N_a+1) * sizeof(char *));

should be: char *seq_a = malloc((N_a+1) * sizeof(char));

If you want to print each char then use %c:

  printf("%c", seq_a[i]);

Upvotes: 1

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