CODEWITHSUNDEEP
ccrashprintf
Shweta
Shweta

Reputation: 5456

printf crashes the program

I am using sprintf to convert int to string and then if I use printf the program crashes otherwise it works fine. Can anyone tell me the reason?

typedef char* string;
buffer[8] = (string*)malloc(sizeof(string));
buffer[8] = sprintf(buffer[8],"%d",inf[i].mPermissions);

It's working fine until here but when I print it like this:

printf("%s",buffer[8]);

...my program crashes.

Upvotes: 0

Views: 5948

Answers (2)

Lundin
Lundin

Reputation: 215115

typedef char* string;

The C language has no string type. A char pointer is not the same thing as an allocated string.

buffer[8]=(string*)malloc(sizeof(string));

Why are you setting item number 9 in buffer to a string? (C is zero-indexed, so 0-8 = 9 items). Is that an array of pointers? Perhaps you meant to allocate a buffer of 8 characters?

It doesn't make any sense to typecast the result from malloc in the C language. In C++ you would have to do a cast.

You are allocating the size of a pointer, not the size of a buffer.

buffer[8]=sprintf(buffer[8],"%d",inf[i].mPermissions);

The first argument to sprintf must be an allocated buffer, not a character (one item of a buffer).

printf("%s",buffer[8]);

You are trying to print a string, but passing an item of a buffer (a character).

To sum this up, I would strongly recommend reading the first chapters of a C language book regarding arrays and pointers before attempting any form of string handling or dynamic memory allocation. If you don't know how something works, don't take a chance at the syntax.

Upvotes: 6

Tom
Tom

Reputation: 44881

My bet is you're not allocating the buffer sprintf is writing into.

You need something like:

int myNumber = 42;
char myBuffer[12];
sprintf(myBuffer, "%d", myNumber);

Upvotes: 1

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