Shift count negative or too big error - correct solution? The thing is that i get 4 compiler warnings on the first four lines with the shift operation: warning C4293: '<<' : shift count negative or too big, undefined behavior I understand why this warning occurs, but i can't seem to figure out how to get rid of it correctly I could do something like: qT |= (unsigned long long)readb() << 56;
Binary shift is too big but isn’t - C++ Forum - C++ Users 1 unsigned long is a 32-bit integer even on some 64-bit platforms 2 2^64 cannot be held by a 64-bit integer Did you mean to 2^63? 3 1 is still of type int You need to convert it to a suitably large type first in order to nor overflow with the shift
Compiler Warning (level 1) C4293 | Microsoft Learn 'operator' : shift count negative or too big, undefined behavior If a shift count is negative or too large, the behavior of the resulting image is undefined Remarks To resolve this issue, you can use a cast on the first operand to expand it to the size of the result type Example The following sample generates C4293, and shows ways to fix it:
How to Fix Shifting by a Negative or Too Large Value: Step-by-Step Gui . . . How to Fix Shifting by a Negative or Too Large Value: Step-by-Step Guide ensure that the shift value is within the permissible range (0 to size of type minus 1) Example: For a 32-bit integer, a valid right shift is from 0 to 31 Example value int shiftAmount = 40; Unintended large shift unsigned result = safeShiftRight(value
c - shift count greater than width of type - Stack Overflow If the size of data_length is 32 bits or less, then shifting right by 56 is too big You can only shift by 0 - 31 Share Improve this answer Follow answered Oct 12, 2018 at 17:38 dbush dbush 228k 25 Also to be able to shift 56 bits the value must be at least 56 57 bits wide Otherwise the behaviour is undefined
Why is there a limit on how many times I can bitwise shift a . . . - Reddit I might dynamically shift it by the result of some calculation, and it's annoying that I have to manually or set to zero if the calculation comes out with a large number Shifting by 8 is the same as shifting by 1 eight times, shifting by 14 is the same as shifting by 1 fourteen times, but when I shift by 434 it's suddenly impossible to guess what I meant and the behavior is undefined?
CCS: Piccolo F28027, shift count too large - TI E2E support forums On a C2800 device the size of int is 16-bits The sub-expression ((0x07 (i+1))<<16) is of type int, so the compiler is correctly warning that the left-shift count of 16 applied to an int is too large and will result in a sub-expression value of zero You need to cast the sub-expression to produce a 32-bit result E g by changing the code to:
Warning C26452 | Microsoft Learn This warning indicates the shift count is negative, or greater than or equal to the number of bits in the shifted operand Either case results in undefined behavior Warning C4293 is a similar check in the Microsoft C++ compiler Code analysis name: SHIFT_COUNT_NEGATIVE_OR_TOO_BIG Example
c - warning: left shift count gt;= width of type - Stack Overflow @whjm: I just referred to the C99 standard and found following: The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros If E1 has an unsigned type, the value of the result is E1 × 2 ** E2, reduced modulo one more than the maximum value representable in the result type If E1 has a signed type and nonnegative value, and E1 × 2 ** E2 is representable in
CCS TMS320F28069: Shift count is too large - TI E2E support forums For my codes, the complier warning me: shift count is too large for the statement (1 << 17), see the figure below Will it hurt my project and how to fix it? over 7 years ago Cancel; 0 Richard Poley over 7 years ago TI__Mastermind 27200 points Yes, it will The compiler treats the '1' as a 16-bit integer, so a left shift of 17 will always