String array iteration causing segfault

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# 1  
Old 08-27-2013
String array iteration causing segfault

I am populating an array of string and print it.
But it going in infinite loop and causing segfault.

char Name[][50] = {

main(int argc, char* argv[])
    int j = 0;

    while(Name[j] != NULL)
            printf("j = %d    Name = %s    len = %d\n", j, Name[j], strlen(Name[j]));



output is
j = 0   Name = yahoo     len = 5
j = 1   Name = rediff len = 6
j = 2   Name = facebook       len = 8
j = 3   Name =     len = 0
j = 4   Name =     len = 0
j = 5   Name =     len = 0
j = 6   Name =     len = 0
j = 7   Name =     len = 0
j = 8   Name =    len = 0
j = 9   Name =    len = 0
12976 Segmentation fault      ./test_code

What is wrong with the code?

Last edited by rupeshkp728; 08-27-2013 at 04:05 PM.. Reason: corrections
# 2  
Old 08-27-2013
You made it an array of arrays, guaranteeing it will never be NULL. array[j] just calculates array + (j*50), which isn't going to be zero.

If you want to store a null pointer, you'll have to store pointers. Then you get address stored at (base_address + j) which is quite capable of being NULL.

char *strings[]={
                NULL };

This User Gave Thanks to Corona688 For This Post:
# 3  
Old 08-27-2013
Thanks coronna for the reply.
Using while(Name[j][0] != NULL) solved my problem.
# 4  
Old 08-27-2013
That means you cannot store an empty string in your array though -- it will be the same as NULL, i.e. first character holding a zero.
This User Gave Thanks to Corona688 For This Post:
# 5  
Old 08-27-2013
yes it seems so and so inspite of being zero loop went infinetely.
# 6  
Old 08-27-2013
I don't understand your question or statement.
# 7  
Old 08-27-2013
I meant whatever you said in your second comment was correct and so inspite of being zero loop went infinetely.
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