Segmentation Fault - C

Question

Why does the following code return with a segmentation fault  When I comment out line 7  the seg fault disappears   int main void         char  s        int ln        puts  Enter String             scanf   s   s         gets s         ln   strlen s      remove this line to end seg fault       char  dyn s    char   malloc  strlen s  1     strlen s  is used here as well but doesn t change outcome       dyn s   s        dyn s strlen s       0         puts dyn s         return 0          Cheers

User · Answer

Your scanf   s   s   is commented out  That means s is uninitialized  so when this line ln   strlen s   executes  you get a seg fault    It always helps to initialize a pointer to NULL  and then test for null before using the pointer

User · Answer

Even better    include  lt stdio h gt  int main void      char  line   NULL    size t count    char  dup line     getline  amp line  amp count  stdin     dup line strdup line      puts dup line      free dup line     free line      return 0

User · Answer

s is an uninitialized pointer  you are writing to a random location in memory   This will invoke undefined behaviour   You need to allocate some memory for s   Also  never use gets  there is no way to prevent it overflowing the memory you allocate   Use fgets instead

User · Answer

Catastrophically bad   int main void         char  s        int ln        puts  Enter String             scanf   s   s         gets s         ln   strlen s      remove this line to end seg fault       char  dyn s    char   malloc  strlen s  1     strlen s  is used here as well but doesn t change outcome       dyn s   s        dyn s strlen s       0         puts dyn s         return 0          Better    include  lt stdio h gt   define BUF SIZE 80  int  main int argc  char  argv            char s BUF SIZE         int ln        puts  Enter String             scanf   s   s         gets s         ln   strlen s      remove this line to end seg fault       char  dyn s    char   malloc  strlen s  1     strlen s  is used here as well but doesn t change outcome       dyn s   s        dyn s strlen s       0         puts dyn s         return 0          Best    include  lt stdio h gt   define BUF SIZE 80  int  main int argc  char  argv            char s BUF SIZE         int ln        puts  Enter String          fgets s  BUF SIZE  stdin      Use fgets  our  cin    NEVER  gets           int ln   strlen s          char  dyn s    char   malloc  ln 1         strcpy  dyn s  s         puts dyn s         return 0

User · Answer

char  s  does not have some memory allocated   You need to allocate it manually in your case   You can do it as follows s    char   malloc 100      This would not lead to segmentation fault error as you will not be refering to an unknown location anymore

[c] Segmentation Fault - C

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