CODEWITHSUNDEEP
cpointerssegmentation-faultprintf
Armin
Armin

Reputation: 107

Segmentation fault using printf when the char pointer (strdup)

I'm new to C-programming. I wrote a function 

void fun (char** argv, int count, int loc);

In my function, I tried to create another array by doing 

char** newray=strdup(argv).

When I tried to execute:

printf(newray[0])

I got a segmentation fault. I fixed it by doing 

printf("%s\n", newray[0]).

However, to output the argv[0], I only need to do this: 

printf(argv[0]).

What is the reason behind it?

Upvotes: 1

Views: 360

Answers (1)

Ishamael
Ishamael

Reputation: 12795

This is because argv is not a string, it is an array of strings, so strdup is not a correct function to copy it. You need to do something along the lines of:

newray = malloc(sizeof(char*) * argc);
for (size_t i = 0; i < argc; ++ i) newray[i] = strdup(argv[i]);

Where argc is the number of elements in argv. You will also need to free all the elements and then the array itself when you are done with it.

Upvotes: 1

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