In-depth analysis on the internal mechanism of converting Char to int/unsigned int

Analysis of converting Char to int/unsigned int: Note: The trial environment is VC ++ 6.0. In VC ++ 6.0, char is regarded as signed Char, so the Integer Range that char can represent is-218 -- + 127 first analyzes the situation of converting to int: 1. If the range indicated by char is between-128 -- + 127, the value size remains unchanged when it is converted to an integer. For example: Char CH = I // I is an integer between-128 and + 127. Int inv = CH; // The value of inv is also an integer 2 between-128 and + 127. If the char expression range is not between-128 and + 127, in this case, take the lowest 8 and convert it to int, for example, char H = 128; int inv = CH; // at this time, the INV value is-128. Why is the result above-128? The analysis is as follows: H content in the memory is (from left to right is from high to low, the same below) 10000000 (in the form of complement code) Put the content in H into Inv, first, judge that the first digit is 1, it must be a negative number, and then convert 10000000 to the original code 10000000, so the INV value is-128. The following is an analysis of the conversion to unsigned INT: 1. If the char type variable value is 0-+ 127, after being converted to the int type, the numeric value remains unchanged. Char H = I // I is an integer between 0 and + 127. Int inv = H // the INV value is also an integer between 0 and + 127. 2. If the char type variable is negative, take 8 lower bits. If 8th is 1, add 1 for expansion. If 8th is 0, add 0 for expansion. For example, char H =-129 unsigned int inv = H. At this time, the INV value is + 127. Why? The analysis is as follows:-129 is written as binary 101111111 (complement representation), and the 8-bit lower value is 01111111. Because the eighth bit is 0, when it is extended to 32int, it is added as 0, so the INV value is + 127 char H =-1; unsigned int inv = h at this time, the INV value is 4294967295, that is, 2 ^ 32-1. Why? The analysis is as follows:-1 is written as binary 11111111 (complement representation), and the 8-bit lower is 11111111. Because the eighth bit is 1, when it is extended to 32int, 1 is added to the high position, therefore, the INV value is reduced by 1 to the power of 2, that is, 4294967295. Well, with the above analysis and examples, we have understood the internal mechanism of converting Char to int/unsigned Int. What about converting unsigned char to int/unsigned Int, I believe that with the above examples and analysis, readers can easily come up with the corresponding conversion mechanism. If you are interested, you can give it a try and I will not go into details here. The end, thanks very much. The final description is as follows: Indicate the author and the source if any reposted. The ownership of this blog belongs to the blog and the author.