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In this example, the value of c1 is multiplied by c2. The product of these values is then divided by the value of c3 (according to operator precedence rules). Assuming that signed char is represented as an eight-bit value, the product of c1 and c2 (300) cannot be represented. Because of integer promotions, however, c1, c2, and c3 are each converted to int, and the overall expression is successfully evaluated. The resulting value is truncated and stored in cresult. Because the final result (75) is in the range of the signed char type, the truncation does not result in lost data.

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Every integer type has an integer conversion rank that determines how conversions are performed. The ranking is based on the concept that each integer type contains at least as many bits as the types ranked below it. The following rules for determining integer conversion rank are defined in C99.:

• No two different signed integer types have the same rank, even if they have the same representation.
• The rank of a signed integer type is greater than the rank of any signed integer type with less precision.
• The rank of long long int is greater than the rank of long int, which is greater than the rank of int, which is greater than the rank of short int, which is greater than the rank of signed char.
• The rank of any unsigned integer type is equal to the rank of the corresponding signed integer type, if any.
• The rank of any standard integer type is greater than the rank of any extended integer type with the same width.
• The rank of char is equal to the rank of signed char and unsigned char.
• The rank of any extended signed integer type relative to another extended signed integer type with the same precision is implementation-defined but still subject to the other rules for determining the integer conversion rank.
• For all integer types T1, T2, and T3, if T1 has greater rank than T2 and T2 has greater rank than T3, then T1 has greater rank than T3.

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In the following example, assume the code is compiled using an implementation with 8-bit char, 32-bit int, and 64-bit long long:

signed char sc = SCHAR_MAX;
unsigned char uc = UCHAR_MAX;
signed long long sll = sc + uc;

Both the signed char sc and the unsigned char uc are subject to integer promotions in this example. Because all values of the original types can be represented as int, both values are automatically converted to int as part of the integer promotions. Further conversions are possible, if the types of these variables are not equivalent as a result of the " usual arithmetic conversions. " The actual addition operation in this case takes place between the two 32-bit int values. This operation is not influenced by the resulting value being stored in a signed long long integer. The 32-bit value resulting from the addition is simply sign-extended to 64 bits after the addition operation has concluded.

Assuming that the precision of signed char is 7 bits and the precision of unsigned char is 8 bits, this operation is perfectly safe. However, if the compiler represents the signed char and unsigned char types using 31- and 32-bit precision (respectively), the variable uc would need to be converted to unsigned int instead of signed int. As a result of the usual arithmetic conversions, the signed int is converted to unsigned and the addition takes place between the two unsigned int values. Also, because uc is equal to UCHAR_MAX, which is equal to UINT_MAX, the addition results in an overflow in this example. The resulting value is then zero-extended to fit into the 64-bit storage allocated by sll.

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