Size of an array in C is smaller than expected - arrays

Question

I have a string in C named buff. It is defined as char *buf = malloc(size);. I'm new to C but think this means that the values of the bytes of data allocated is stored in buf. All code is below
When I run sizeof(buf) it gives a size of 8 despite it clearly being full of more than 8 characters. I also have problems with taking the returned pointer and putting it through strlen which gives a value double what I expect.
This is a function to create an array of characaters read from a file.
The comments below used an example file of input.txt with contents ()()(()()(
const char* readFile()
{
FILE* fp;
int c;
int nch = 0;
int size = 100;
fp = fopen("./input.txt","r");
//allocates a block of 100 bytes and sets buf as the values associated with those values
char *buf = malloc(size);
if(buf == NULL)
{
fprintf(stderr, "out of memory\n");
return NULL;
}
while((c = fgetc(fp)) != EOF) //sets c to the next character of fp and checks it isn't the end
{
if(nch >= size-1) //size-1 as need an extra character for '\0'
{
/* time to make it bigger */
//printf("size: %d \n",size);
size += 50;
buf = realloc(buf, size);
//size_t n = arraySize(buf);
//printf("buf is %zu chars long \n",n);
if(buf == NULL)
{
fprintf(stderr, "out of memory\n");
return NULL;
}
}
buf[nch++] = c; //steps nch and stores c
}
buf[nch++] = '\0';
printf("size: %d\n",size); //output 100
printf("nch: %d\n",nch); //output 11
buf = realloc(buf, nch); //sets the buf to the size it actually needs
printf("buf is %d long\n",sizeof(buf)); //outputs sizeof(buf) as 8.
printf("buf[0] is %d long\n",sizeof(buf[0])); //outputs sizeof(buf[0]) as 1.
printf("\"%s\"\n", buf); //shows it reads the file properly. outputs "0123456789"
return buf;
}
int main(){
const char *fileContents = readFile(); //defines it as the address of the first character
printf("\"%s\"\n", fileContents); //outputs the file contents correctly
int count = 0;
printf("strlen: %d\n",strlen(fileContents)); //outputs 10
int i, lenStr = strlen(fileContents);
for(i = 0; i < lenStr; i++){
char c = *(fileContents+i); //sets character at position i to c
printf("character: %c i: %d\n",c,i); //outputs expected value
switch(c){ //this works sort of and correctly identifies values
case '(':
printf("increasing count\n");
count++;
break;
case ')':
printf("decreasing count\n");
count--;
break;
default:
printf("things are going wrong \n");
break;
}
fileContents++;
}
printf("go up %d floors",count);
getchar();
return 0;
}
output of for loop with select statment
character: ( i: 0
increasing count
character: ( i: 1
increasing count
character: ( i: 2
increasing count
character: ) i: 3
decreasing count
character: ) i: 4
decreasing count
character: i: 5
things are going wrong
character: i: 6
things are going wrong
character: i: 7
things are going wrong
character: e i: 8
things are going wrong
character: s i: 9
things are going wrong
For some reason the output is in the wrong order before it starts breaking.
It feels like the problem is I don't fully understand pointers. I'm just not sure how I am misunderstanding them.

Answer 1

The main problem is that you are doing fileContents++;. This adds one to the pointer fileContents, sliding the place to view.
You should remove this so that the characters will be processed one-by-one thanks to i++ in the for loop.
Another problem is that you are invoking undefined behavior by passing data having wrong type to printf. sizeof operator and strlen returns size_t, but %d expects int. The correct format specifier to print size_t is %zu.