I wanted to know what my maximum value could have time_t, so I wrote a small program to help me. He needs one argument: the number of bytes (1 byte = 8 bits). So I wrote it and tested it. It works great for all values from 1 to 4, but in 5 and above it also edits the "signed" bit (I don't know what it's called). Can someone explain:
#include <stdio.h>
int main(int argc, const char **argv) {
if(argc != 2) {
fprintf(stderr, "Usage: %s bits/8\n", argv[0]);
return -1;
}
unsigned int bytes;
sscanf(argv[1], "%u", &bytes);
unsigned int i;
signed long long someInt = 0;
size_t max = bytes*8-1;
for(i = 0; i < max; i++) {
someInt |= 1 << i;
}
max++;
for(i = 0; i < max; i++) {
int isAct = (someInt >> max-i-1) & 1;
printf("%d", isAct);
if((i+1) % 8 == 0) {
printf(" ");
}
}
printf("\n");
printf("Maximum size of a number with %u bytes of 8 btis: %lld\n", bytes, (long long)someInt);
return 0;
}
My testing:
Script started on Sun Jan 30 16:34:38 2011
bash-3.2$ ./a.out 1
01111111
Maximum size of a number with 1 bytes of 8 btis: 127
bash-3.2$ ./a.out 2
01111111 11111111
Maximum size of a number with 2 bytes of 8 btis: 32767
bash-3.2$ ./a.out 4
01111111 11111111 11111111 11111111
Maximum size of a number with 4 bytes of 8 btis: 2147483647
bash-3.2$ ./a.out 5
11111111 11111111 11111111 11111111 11111111
Maximum size of a number with 5 bytes of 8 btis: -1
bash-3.2$ ./a.out 8
11111111 11111111 11111111 11111111 11111111 11111111 11111111 11111111
Maximum size of a number with 8 bytes of 8 btis: -1
bash-3.2$ exit
exit
Script done on Sun Jan 30 16:35:06 2011
I hope to learn from this, so I would really appreciate it if someone takes the time to take a look at this.
ief2
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